# Challenge: A Cool Double Summation + Integral Combo

Share
Embed
• Published on Aug 9, 2018
• Congratulations to Gabriel N., Allaizn, Benjamin Wang, Essentials of Math, iQuickdraw X, staffehn, Quwertyn, NoName, Hung Hin Sun, and Jacob Glidewell for successfully solving the last week's math challenge question! Gabriel N. was the first person to solve the question.
Your support is truly a huge encouragement.
Every subscriber and every like are wholeheartedly appreciated.
Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.

• traso 10 months ago

Can you please solve this for me please x^x=x•10^x ive been trying to find the value of x without sucess it is around 12 but i cant find the exact number :(

• Kevin Tong 10 months ago

How come I get the notification for this 5 days after it's posted.....

• Debraj Banerjee 10 months ago +1

This is not a challenge . it's a simple math problem with unity answer

• Benjamin Wang 10 months ago +1

LetsSolveMathProblems Please cover all these relevant solution methods (or as many as you can) that viewers proposed. 1. Integration directly, 2. switching sums to simplify;3 Gamma .4 Laplace. Problems with more solutions are all the more fascinating and profound

• Benjamin Wang 10 months ago

LetsSolveMathProblems thank you.”:) your videos are great- thorough and instructive while keeping a right pace

• LetsSolveMathProblems  10 months ago +1

I will cover the first two solutions for sure. I am debating on how much, if any, of the last two solutions to cover. After all, Laplace transform for this particular case and properties of gamma function are both derived from direct integration. I may briefly mention them, but I probably would not explain them very in-depth.

• Mr Sharingan 10 months ago

1

• Dishara Hettiarachchi 10 months ago

the integration of x^b.e^-ax is equal to (b!)/a^(b+1) say equal to Y
hence Y/(b!) equals to 1/a.a^b
the sum of geometric series 1/a.a^b where b=1 to b=infinity is 1/a(a-1)
now the n th partial sum of series 1/a.(a-1) where a=2 to a=infinity should calculate.
notice that 1/a(a-1)=1/(a-1) - 1/a
put 1/a=f(a)
then 1/a(a-1)=f(a-1) - f(a)=U(a)
since U(2)=f(1)-f(2)
U(3)=f(2)-f(3)
U(4)=f(3)-f(4)
.
.
.
U(n)=f(n-1)-f(n)
then by adding them n th partial sum of series 1/a.(a-1)= f(1)-f(n)=1-1/(a-n)
now notice that lim n-->infinity [1-1/(a-n)]=1

• Fady Omari 10 months ago

We can switch the second summation and the integral, the sum from 1 to infinity of x^b/b! is e^x-1 , distribute this with 1/e^(ax) then we integrate ,
We have the first sum of (e^[x(1-a)]/(1-a) + e^(-ax)/a ) with the bounds of integration
If we plug infinity we get 0 ( e^(inf*(1-a)) since a is positive so we have e^-inf which is zero )
So at the end:
Summation from a=2 to infinity of 1/(a-1) -1/a is 1 by telescoping series test .
So we get 1

• dominofan238 10 months ago

The substitution u=ax lets us turn the integral into Γ(b+1)*1/b! * 1/(a^(b+1))=1/(a^(b+1)). The sum over b is a geometric series with r=1/a

• adandap 10 months ago +2

The answer is one. Do the integral first by substituting t = ax, and it gives Gamma(b+1)/[a^(b+1)b!] = 1/a^(b+1). Then do the b summation - the series is 1/a^2 + 1/a^3+... = 1/[a (a-1)] since it is a geometric series. Then partial fractions gives the sum over a of 1/(a-1) - 1/a, which telescopes, to leave 1. Ta da!

• Federico Espil 10 months ago +1

No onda will believe me but I have just calculated this mentally it's 1 I can't believe it. The trick to avoid such a hard integral I thought to flip e^(-ax) so I thought the integral as laplace transform of x^b Anyway I feel like terrific monster I did partial fractions by cover up all in my head I can't believe I got the right answer congratulations to my self lol I swear it's true

• Benjamin Wang 10 months ago

Federico Espil @LetsSolveMathProblems perhaps you should cover all these possible solutions in your solution video? This is fascinating. Direct integration, switching sums, Gamma function, and now Laplace transform??

• VOLC4NICxSP4RK 10 months ago

How do you even start

• higher mathematics 10 months ago

answer is 1, first with the integral, bring the factorial out as a constant then the integral turns into (0,inf) of x^b*e^-ax by using the substitution u=ax du=adx then with some simplification the constants out front read 1/((b!)(a^(b+1)) followed by the integral (0,inf) u^b*e^-u which is gamma(b+1)=b! in which cancels with b! in front of the integral so you're done with the integral and left with the sum from b=1 to inf of 1/(a^(b+1)) which you can turn into a geometric series 1/(a(a-1)) so you now have the final sum of a=2 to inf of 1/(a(a-1)) in which you can turn into a limit of its partial sum which looks like lim a-->inf (1-(1/n)) in which the 1/n goes to zero and you are left with just 1.... :>

• Jean le Ronde d'Amelbert 10 months ago +1

The answer is 1. You have to rearrange the sums and integral so that the integral is outside (I kind of just assumed that this works). Then you can easily see that the nested sum can be turned into a product of sums. One summation is simply the Taylor expansion of e^x minus 1 while the other sum is simply a geometric series. The summations simplify to e^-x and the integral of that is simple enough.

• Marco Dalla Gasperina 10 months ago

I get -1. The inner sum is e^x - 1 (because b starts at 1). in the end you get the sum a=[2,inf] 1/a * 1/(1-a) which is -1.

• Woof 10 months ago

The answer is evaluate to 1.
First factor 1/(b! ) out from the integral, leaving int_0->inf x^b e^(-ax) dx which is very close to gamma distribution function
The integral is evaluated to gamma(b+1)/a^b
gamma(b+1) is canceled out with 1/(b!), leaving ΣΣa^-b.
Σa^-b is evaluated to 1/(a(a-1)), with the help of partial fraction, rewrite to 1/(a-1) - 1/a
And finally, Σ(1/(a-1) - 1/a) is a telescoping series which evaluate to 1.

• Just Weird 10 months ago

This is my first time trying these questions and I don't study maths so I hope I will understand enough :)
I tried starting to actually do this integral, well this is outside my scope. I tried looking at the definition for the laplace series and just tried it with ax=st and it worked for that part. the first summation wasn't to difficult.But I for some reason don't understand the last summation. I tried to solve it using u_1* (1-r)^inf/(1-r) but I do not get 1. Is there a trick I should see for this last summation?
*edit: I just filled out the summation and noticed I forgot the very first term (+1) and that solved it. Well I am happy I was able to solve it.

• Fares BERARMA 10 months ago +1

First integrate from 0 to inf x^b*exp(-ax) wich is equal to b!/a^(b+1)
Somme became from a=2 to inf 1/a^2+1/a^3+...1/a^n as n goes to inf the result will be somme from a=2 to inf (1/(a-1)-1/a)=1-1/n=1

• Bruno Alejandro Andrades 10 months ago

First we analize the inside sum, if we switch the integral by the sum u get, int((1/e^ax)sum x^b/b!), since b starts at 1, int(( 1/e^ax)(e^x-1)), then u split the fractions, add up the exponents and integrate the usual exponential function, taking the limit to calculate the boundaries, u get a lot of e^-(inf), and end up with sum(1/(a-1) - 1/a), u split this into two sums, for the first one we can say it's equal to sum from a =2 to inf of 1/a and then u add 1/1, this comes from changing the index, since taking a sum to infty is technocally taking the limit, we don't care about the convergence of the sums and just substract one from another, so at the end we get an answer of 1

• ba ke 10 months ago +1

This is 1.
Partial integration should be repeatedly applied.
Then it is better to sum it by partial fractional decomposition with the formula of the sum of the geometric progression.

• olivier2000497 10 months ago +1

First reminder: sum from n=0 to infinity of (x^n)/n! is equal to exp(x)
To begin with, I switch the integral and the sum for b. Then I get 1/exp(ax) out of the sum, since it doen't have any b. We have something that looks like the first reminder, except b begins with 1. To solve this, we just add 1 and subtract 1 because for b=0, (x^b)/b! = 1. We end up with (exp(x) - 1)/exp(ax) in the integral.
Again, we switch the integral and the second sum. We get (exp(x) - 1) out of the sum, and we have a geometric serie with q = 1/exp(ax) = (1/exp(x))^a and |q| < 1 (q is positive). By using the geometric serie formula, we have sum from a=2 to infinity of 1/exp(ax) equal to (1/exp(2x))/(1 - 1/exp(x)) = 1/(exp(x)*(exp(x) - 1)). Surprise ! We have (exp(x) - 1 on top and bottom of the fraction !
By simplifying, we get integral from 0 to infinity of 1/exp(x). It's easy to integrate since 1/exp(x) = exp(-x) and we have 1.

• Zeick87 10 months ago

A scan of my solution is here: imgur.com/a/1h0pSSb
I performed the integral by substituting y=ax, getting Euler gamma function. The resulting b-summation is a geometric series, which converges since the magnitude of the absolute value of the division of two consecutive terms in the series are less than one. The final a-summation is broken apart by partial fraction decomposition and the following cancellation of terms in the sum leaves only first term remaining, giving the final answer, 1.

• end my misery 10 months ago +1

first time i got these questions right!
at first i swapped the integral and summation(the b one), after that i pulled the 1/e^(ax) out of the summation, and then we were left with the sum from a=2 to inf of the integral from 0 to inf of 1/e^(ax) times the sum from b=1 to inf of x^b/b!, and that sum is exactly e^x -1 !(exclamationmark) since the taylor series of e^x starts from b=0.
now we're left with the sum from a=2 to inf of the integral from 0 to inf of e^-(ax) * (e^x-1). opening brackets leaves us with-
the sum from a=2 to inf of the integral from 0 to inf of e^(1-a)x - e^-(ax), which is equal to(after integrating) the sum from a=2 to inf of [e^(1-a)x/(1-a) + e^(-ax)/a] from 0 to inf.
the infinity terms vanish, since a>2 its specifically greater than 1, so all of the e exponents are negative, hence e^-inf=0. plugging 0 to the terms leave us with
the sum from a=2 to inf of -1/(1-a) - 1/(a). now that's a telescoping sum
plugging is values lets us see that this is indeed a telescoping sum, and we're left with 1 + the lim as a approaches inf of 1/a, which is simple 1!
im actually so excited i got this question right. please correct me if i did anything wrong! only thing i think is suspecious is the fact that i interchanged the summation and integration operators, but since it converges(i think) it should work. i never dealt with any question of the sort..

• Suneet Iyer 10 months ago +1

The improper integral evaluates to 1/(a)^b+1
The first summation is then reduced to an infinite geometric series which evaluates to 1/a(a-1). Expressing as a partial fraction yields a telescoping series evaluating to 1

• Rhombic Dodecahedron 10 months ago

The answer is 1. If you use the substitution u=ax, the integral becomes (1/a)^(b+1) * 1/b! * Γ(b+1), where Γ(n) is the gamma function which has the property Γ(n+1) = n!, so Γ(b+1) = b! which cancels with the 1/b! so that you get (1/a)^(b+1). For each a, the sum of b from 1 to ∞ becomes a geometric series with starting term 1/a^2 and factor 1/a, using the formula first term/(1-r) it becomes 1/(a*(a-1)). Using partial fraction decomposition, the infinite sum of a becomes a telescoping series which results in the answer 1.

• Tobias Guivarch 10 months ago +1

First we can evaluate the integral, indeed it's the gamma fonction ( with a substitution ax=t) . Then we have a geometric serie that converge. Finally we have to evaluate the infinite sum of 1/a(a-1) that we can transform with the substitution k=a-1 and with partial fraction and telescopic sum we end up with 1- Lim(n->inf) 1/(n+1) which equal to 1.

• Benjamin Wang 10 months ago

Tobias Guivarch FYI the host apparently only accepts unedited nonreply comments (so that the order of first solutions will be correct)

• Shubham Dewangan 10 months ago

Infinity

• nkoukou 10 months ago +1

Switch order, which we can do because sums are convergent. Sum over b is the Taylor expansion of (e^x - 1), while sum over a is an infinite geometric series evaluating to (e^-x / [e^x - 1]), so we are left with e^-x as the integrand.

• BARRY Mamadou Saïdou 10 months ago

Hello there.
Using partial integration we can easily show that the value of the integral is b*Gamma(b)/(b!*a^(b+1)). Where Gamma is the famous Euler function.Knowing that Gamma(b)=(b-1)! for every integer b greater than zero, we can therefore simplify the previous result as follows: 1/(a^(b+1)).
What is left now is the double sum. The sum of (1/a)^b for b=1 to infinity while a remains fix and greater or equal to 2 is geometric and equal to 1/(a-1). Then the infinite sum becomes: sum(1/a*1/(a-1); for a=2 to infinity); this is a telescopic sum than can be easily evaluated: let's call N an integer greater than 2;
Sum(1/(a-1)-1/a; for a=2 to N) is equal to 1-1/N. Therefore if N approaches infinity the sum will be equal to 1. This is the result of the question.

• Kyle Ramsey 10 months ago +2

Answer is 1. Switch the second summation sign with the integral sign, then allow the first bound of the summation sign to be b = 0 by subtracting the case when b = 0. The power series of the exponental function reveals itself in the first integral, so both integrals are trivial to evaluate. We are then left with an infinite series, and if we look at its partial sums we realize we can use induction to prove that the limit of the partial sums is simply 1.

• Mehdi Nazerian 10 months ago +1

Hi:answer is "1" by gamma function, the solution of improper integral is "a^(-(b+1)) cosequently "the first summation is "(a/(a-1))" and eventually the last summation equals to "1" by nested series.
Thank you so much👏👏👏 for very very nice video .peace be upon you.* I am Mehdi 'math teacher' from Iran*🙏🙏🙏

• Mehdi Nazerian 10 months ago

Benjamin Wang 🙏🙏❤❤⚘⚘

• Benjamin Wang 10 months ago

Seems like only you solved this with Gamma function. Would you take the time to elaborate please? Thank u

• Sir Lanzelot 10 months ago +1

Invoke some fancy convergence theorem, then switch integration and summation, then use the powerseries representation of the exponential function and then the geometric series, after that you have only e^(-x) left to integrate from zero to infinity which equals 1.

• Phoenix Fire 10 months ago

I need help. I'm trying to calculate the infinite sum of ((b-n)!/b!) starting at n. This isn't for the problem; I just got to thinking about it at the last step to the problem.
Obviously, n>1.
Individually, I can break down each sum using 1/(xy)=(1+x)/(xy)-1/y, 1/(xy)=(1-x)/(xy)+1/y and a little smarts, but I can't tackle the whole thing.
Examples:
n=2, infinite sum from 1 of 1/(a(a+1))=1/a-1/(a+1) terms cancel leaving only the first and last which =1-0=1
This one was a trivial application of the algebraic property I noted above.
n=3, infinite sum from 1 of 1/(a(a+1)(a+2))=(1/a-1/(a+2))/2-1/(a+1)+1/(a+2) =(1+1/2)/2-1/2
This one was a fun application of the algebraic properties I noted above.
From here I noticed that there's a simpler property (n-m)/((a+m)(a+n))=1/(a+m)-1/(a+n)
Solving each individual sum is kind of simple now. I did n=4 and did a rough calc for n=5. I'm sure that if I did a few more examples, I would see the pattern completely, but I'm tired so I'll settle for the insight I gained at n=5.

• The ans is 1.
First evaluate the summation of b . Get (e^x-1)/(e^ax) in the integral. Then evaluate the integral, get 1/(a-1) - 1/a, lastly expand the summation and get 1.

• Daniel Flores 10 months ago +1

The solution is 1, use tabular integration to evaluate the integral, every term except the last gets canceled 1/(a^(b+1)).
then sum the geometric series to get 1/a(a-1) then separate the fraction into 1/(a-1)-1/a and realize that it's a telescoping series in which the first term is the only one not eliminated.
You could just switch the order but I fear that would require justification beforehand, the math part of my brain keeps grumbling something to me about "uniform convergence" but im too lazy to remeber that stuff right now.

• Kyle Noble 10 months ago +7

First we note that the integral contains an e^(x times a constant) and goes from 0 to infinity.
The first thing I thought of was the definition of the Laplace transform:
integral from 0 to inf of f(t)*e^(-st) dt
Since 1/b! is a constant linear factor, it can be moved outside the integral leaving:
integral from 0 to inf of e^(-ax)*x^b dx
If we apply the definition of the Laplace transform, we take a as our s, x as our t, and t^b (previously x^b) as our f(t).
In other words, we're left with the L{x^b} as a function of a.
This is a standard, straight-forward transform which is commonly derived.
L{x^b}(a) = b!/(a^(b+1))
Now the integral is gone and we're left with the sum from a=2 to inf of the sum of b=1 to inf of (1/b!)*b!/(a^(b+1))
The b! terms cancel and we have an infinite geometric series.
sum from b=1 to inf of 1/(a^(b+1))
At b=1 we have 1/a^2; at b=2, 1/a^3; b=3, 1/a^4; etc.
The first term is 1/a^2 and the common ratio is 1/a. (-1inf) -1/a = 1/1-0=1.

• Rohan Shinde 10 months ago +1

The integral is the standard gamma function after substitution t=ax. When turning towards summation of "b" we have a geometric progression arriving there and at last for the summation with "a" as variable, it's just the partial fraction decomposition and ending up with the telescoping series

• Rohan Shinde 10 months ago

Daniel Flores Thanks

• Daniel Flores 10 months ago

That observation of the gamma function is a nice catch and deserves a mention in the following video.

• nin10dorox 10 months ago

1

• bridogg154 10 months ago

I feel like a zeta function is close by...

• Zbigniew Brzezinski 10 months ago

Answer is 1. Surely the easiest way is to use Gamma function. We can get an integral function of (ax)^b*e^(-ax)d(ax), then we get b!. Now, 1/a^(b+1) is a geometric series, and it is convergent with a>=2, and we get 1/(a*(a-1)) by using infinite g.p. formula, then split it into 2 terms: 1/(a-1) - 1/a.

• Rohit Chaurasiya 10 months ago

Ans is 1 .
First we solve the integral using Gamma function and we get 1/(a)^b+1 which is a g.p. And then we solve the Sigma by using infinite gp formula .we get the series 1/(a)(a-1) which can be written as. 1/(a-1)-1/a and sovling the sum after cancellations we get 1-1/a where a tends to infinity. Therefore we get ans.1

• Erin Cobb 10 months ago

1
SUM_a=2^Inf (SUM_b=1^Inf (Int_0^Inf (x^b/((e^ax)(b!)))dx))
From Integration by parts, x^b -> bx^(b-1) -> b(b-1)x^(b-2) -> ... -> (b!)x^0 & e^-ax - > e^-ax/-a -> e^-ax/a^2 -> ... -> e^-ax/(-a)^(b+1) & answers alternate by +/-/+/-, so that all terms are negative.
The integral from 0->Inf, the e^-ax dominates when x->Inf, so all of those evaluate to 0. Only the constant value from the Integration by parts remains so you get -(b!)/((a^(b+1))(b!)) = 1/a^(b+1)
That's a geometric series, so you get left with SUM_a=2^Inf( (1/a^2)/(1-(1/a)) ) = SUM_a=2^Inf( 1/(a^2-a) ) = SUM_a=2^Inf( 1/a(a-1) )
1=A(a-1)+Ba A=-1, B=1 -> SUM_a=2^Inf( (1/(a-1)) - (1/a) ) = lim_n->Inf (1-1/2) + (1/2-1/3) + (1/3 - 1/4) + ... + (1/(n-2) - 1/(n-1)) + (1/(n-1) - 1/n) =lim_n->Inf 1+(-1/2+1/2)+(-1/3+1/3)+...+(-1/(n-1)+1/(n-1)-1/n = lim_n->Inf 1-1/n = 1

• Riven Mercy 10 months ago

1

• Ethan Winters 10 months ago +4

1
We can use the linearity of the integral and the absolute convergence of the series to move the summations to the inside of the integral. Now we only have to focus on the sums. Since the expression x^b/(e^ax)*(b!) can be split up into two factors, one in terms of only a (1/e^ax), and the other in terms of b (x^b/b!) we can use the distributive property twice to separate them. Now can evaluate each sum separately and multiply the results. The first sum from 1 to inf of x^b/b! is simply e^x - 1 by the taylor series definition of e^x. The second sum, from 2 to inf of 1/e^ax, is a simply geometric series, and converges to e^-x/(e^x - 1) for x > 0. Multiplying these two expressions together gives us e^-x. Integrating this from 0 to inf results in the final answer of 1.

• Yousef Romiah 10 months ago +1

No need to change order. Integrating gives b!/a^(b+1). The b sumnation is a convergent geometric series that gives 1/a(a-1) which is a telescoping series of values 1.

• M. Shebl 10 months ago

Yousef Romiah i know right, i.t was 2:00 AM for me solving the problem and after posting later i realized there was no need to do that

• 35cut 10 months ago

Yousef Romiah
How did you integrate x^b / e^ax ?
Switching the sum and the integral was the first thing I saw and didn't even think about integrating that mess on the inside.

• Ghylherme Patriota 10 months ago

First the integral will be exp (-ax) x^b / b !. By doing a u-substitution the resulting integral will be 1 / (a ​​^ (b + 1)) times the gamma function of (b + 1). O b! resultant is annulled with b! of the denominator, with only the double sum of 1 / (a ​​^ (b + 1)) remaining). With a change of n = (b + 1), the resulting series will be the geometric series of (1 / a)^n with initial n equal to 2. The final result will be the sum of 1/(a (a ​​-1) ). Applying partial fractions will have the sum of 1/(a-1) - sum of 1/ a with. "a" starting at 2. This will result in 1 - 1/2 + 1/2 -1/3 + 1/3 .... + 1/n - 1/(n + 1), where the only remaining terms will be 1 - 1 / (n + 1). Applying the lim n -> infinite 1 - 1/(n + 1) = 1.

• Minh Cong Nguyen 10 months ago +1

First we can switch the integral with the sum over b, and use the expansion of e^x=1+x+x^2/2!+x^3/3!+... therefore the initial expression is equal to sum of integral of (e^x-1)dx/e^(ax) with a = 0..infinity.
Then again, we change the integral with the sum over a to get integral of (e^x-1)(1/e^2x + 1/e^3x+....)dx
calculating the geometric serirs give us integral of 1/e^x dx from x=0 to infinity, which is equal to -e^-x from 0 to infinity, which is equal to 1.

• The answer is 1 I used telescopic series and the expation of e^x by taylor😊 and obviously a very easy integration

• Felix Trihardjo 10 months ago

Substitute u = ax and pulling out the 1/b! constant from the integral, the integral becomes b!/a^(b+1). The factorial cancels out and you are left with summation of 1/a^(b+1) where b >= 1 and a >= 2. Since -1 < 1/a < 1, the first summation becomes 1/(1-1/a)-1-1/a. Simplifying you get 1/(a-1)-1/a where a >= 2 which is a telescoping sequence which evaluates to 1.

• Benoit Depeault 10 months ago

I think the answer is 1

• Benjamin Wang 10 months ago

Seems correct. However, just FYI the host only accepts unedited not-reply comments

• Benoit Depeault 10 months ago

The path I followed is different than what a read after getting 1.
I started by developping the first part of the b summation getting sum(a=2 to infinity)*(integral from 0 to infinity with b=1 + integral from 0 to infinity with b=2...
Solving for the first term only, so sum(a=2 to infinity)*integral of 0 to infinity with b=1 I got 1/2^2+1/3^2+1/4^2...
Then I did the same for the second term, so sum(a=2 to infinity)*integral from 0 to infinity with b=2 and I got 1/2^3+1/3^3+1/4^3...
We can in fact prove that the logic applies for the other terms (long to write in comment so I will let you do this short proof) so with the next one we will get 1/2^4+1/3^4+1/4^4 ...
We can then add these results (to complete the a sum) and write them in the form (1/2^2+1/2^3+1/2^4...)+(1/3^2+1/3^3+1/3^4...)+... We now have a sum of many geometric series(or almost) we just have to evaluate them as if the geometric sums were complete and then substract to each of them the first two terms of the sum.
Evaluating them this way we get 1/2+1/6+1/12+1/20+1/30+1/42... We can then define Sn as the sum of the n-th first terms of this sum. S1=1/2, S2=2/3, S3=3/4 and so Sn=n/n+1. To get the answer we take lim(n to infinity)Sn=lim(n to infinity)n/n+1 and by the hospital rule we get lim(n to infinity)1=1 final answer.

• Aryan Jain 10 months ago

Prove it

• N J 10 months ago

The answer is one. The Integral Term is just the Laplace Transform of x^b/b! at a, that is, L{x^b/b!}(a), which evaluates to a^(-(b+1)). The sum over b from 1 to infinity is thus a geometric series with the first term being one, thus we have to subtract one from the sum, which gives us the sum of a from 2 to infinity of 1/a*(1/(1-1/a)-1). This sum is a telescoping series, as all terms but the 1 in the very first term vanish, thus we have proven that the expression given in the video is equal to one.
PS: I'm sorry for the lackluster explanation, but it's 3AM here :P

• Felipe Lorenzzon 10 months ago

Sum from a=2 to inf of {sum from b=1 to inf of{1/b! • integral from 0 to inf of (x^b)e^(-ax)dx}}
Use u=ax, so that du/a=dx
Sum from a=2 to inf of {sum from b=1 to inf of{1/((a^(b+1))(b!) )• integral from 0 to inf of (u^b)e^(-u)du}}
We now can recognize the integral as the pi function (the gamma function shifted to coincide with the factorial of integers) of b, i.e., b!
Sum from a=2 to inf of {sum from b=1 to inf of{b!/((a^(b+1))(b!) )}}
Sum from a=2 to inf of {sum from b=1 to inf of{1/(a^(b+1)}}
We can expand the second sum:
Sum from a=2 to inf of {-1/a+1/a+1/a^2+1/a^3+...}
We can recognize that as a geometric series and use the usual method of setting it equal to, let's say, s and then multiply everything by *-a*. Add s and -as, and isolate s. The right-hand side has a bunch of nice cancelations. I won't write all that down here because it would be awfull and messy.
Plugging in the result of the geometric series:
Sum from a=2 to inf of {-1/a+1/(a-1)}
Sum from a=2 to inf of {1/(a-1)-1/a}
expanding this sum:
1/1-1/2+1/2-1/3+1/3-1/4+...
This is a telescoping series, where the last term is zero becaus limit as a gos to inf of 1/(a-1) is 0
the only term we are left with is the first: 1.
So, our final answer is 1

• staffehn 10 months ago +3

This is a great exercise to all calculus lovers (unlike me xD).
In the end, I did most steps the way Hung Hin Sun did, except I didn’t see how to treat "1/[a(a-1)]" in an elegant way (so I showed it’s partial sum being n-1/n by mere induction). I am also pretty impressed at how effortless WolframAlpha handles this whole problem (you can literally type it in and get the correct answer, 1) and it also handles some of the sub-terms for error-checking.
Anyways, as I like details and I’m not going to be first anyways, I’ll do a little writeup of my notes now:
First we’ll consider int_0^inf(x^b/(e^(ax)*(b!))). Linearity allows the 1/(b!) constant factor to be but aside, and int_0^inf(x^b/(e^(ax)) will be tackeled for all integer b≥1 inductively by partial integration. As we know, int(f g') = [f g] - int(f' g). We’ll use f=x^b, g'=1/e^(ax)=e^(-ax). Then f'=bx^(b-1), g=(1/(-a))e^(-ax). Then [f g]_0^inf = b/-a [x^b/e^(ax)]_0^inf = 0 - 0, as long as a > 0. Thus int_0^inf(x^b/(e^(ax)) = int(f g') = - int(f' g) = -int_0^inf(bx^(b-1)*(1/(-a))e^(-ax)) = b/(-a) * (-int_0^inf(x^(b-1)/e^(ax))) = b/a * int_0^inf(x^(b-1)/e^(ax)). This continues ... = b/a * (b-1)/a * int_0^inf(x^(b-2)/e^(ax)) = b/a * (b-1)/a * (b-2)/a * int_0^inf(x^(b-3)/e^(ax)) = ... = b!/a^b * int_0^inf(x^0/e^(ax)), inductively until b reaches 0. Then int_0^inf(x^0/e^(ax)) = int_0^inf(1/e^(ax)) = int_0^inf(e^(-ax)) = 1/(-a) * [e^(-ax)]_0^inf = 1/(-a) * (0 - 1) = 1/a. All in all then int_0^inf(x^b/(e^(ax)*(b!))) = 1/b! * int_0^inf(x^b/(e^(ax)) = 1/b! * b!/a^b * int_0^inf(x^0/e^(ax)) = 1/b! * b!/a^b * 1/a = 1/a^(b+1).
Then Σ_b=1^inf (int_0^inf(x^b/(e^(ax)*(b!)))) = Σ_b=1^inf (1/a^(b+1)) = Σ_b=0^inf (1/a^(b+2)) = 1/a² * Σ_b=0^inf ((1/a)^b) =(a geometric series)= 1/a² * 1/(1 - (1/a)) = 1/(a²-a).
I got stuck (as mentioned above, didn’t find an elegant way) on summation of 1/(a²-a) next, but WolframAlpha suggested: Σ_a=2^N (1/(a²-a)) = (N-1)/N. This is indeed true, shown easily by induction, where N=2 is simply: 1/(2²-2) = 1/2 = (2-1)/2, and N->N+1 goes by Σ_a=2^N+1 (1/(a²-a)) = Σ_a=2^N (1/(a²-a)) + (1/((N+1)²-(N+1))) = (N-1)/N + 1/(N² + 2N + 1 - N - 1) = (N-1)/N + 1/N(N+1) = (N-1)(N+1)/N(N+1) + 1/N(N+1) = (N²-1 + 1)/N(N+1) = N²/N(N+1) = N/N+1.
Finally, Σ_a=2^inf (Σ_b=1^inf (int_0^inf(x^b/(e^(ax)*(b!))))) = Σ_a=2^inf (1/(a²-a)) = lim_N->inf ( Σ_a=2^N (1/(a²-a)) ) = lim N->inf ( (N-1)/N ) = 1.

• staffehn 10 months ago

thanks, I knew that already, it was exactly this approach I meant by referencing "Hung Hin Sun"s comment which was I think the only solution already present, when I started turning my handwritten notes into this comment. They used exactly this telescoping sum in their comment. I still did it in the comment the way I did because that’s the solution I came up with myself without copying anyone else’s ideas (except from ideas WolframAlpha gave me xD).

• Arsène Ferrière 10 months ago

An easy way to deal with sum_(a=2)^∞ 1/(a(a-1)) is by decomposing 1/(a(a-1)) as 1/(a-1)-1/a this way you create a telescoping sum : sum_(a=2)^N 1/(a(a-1))=sum_(a=2)^N 1/(a-1)-1/a =1-1/N (first term-last term) when N goes to ∞, the only remaining term is the 1

• staffehn 10 months ago +1

And now in the mean time ppl started switching sums and integrals, great xDD. Good night, it’s late here, but I like these other solution as a motivation for me to repeat some contents of the calculus lectures I attended a while back to find the details on those kind of theorem(s) again.

• Foster Sabatino 10 months ago +2

Answer is 1. Wen can switch the first integral and summation since the each integrand is a positive sequence of integrable functions.This leads to the taylor expansion of e^x-1. So, we have the summation from a=2 to infinity of the integral from 0 to infinity of e^((1-a)x)-e^-(ax)dx. Evaluating this integral from 0 to infinity lead to 1/a-1/(a-1). The sum is from a = 2 to infinity of this telescoping series, evaluating the first term gives us the answer to this sum, which is 1.

• Cesare Angeli 10 months ago +3

We can change the sum over b with the integral. In this way we get the exp (x) from the term in b.
Then we remain with elementary integrals of exponential we can analitically solve obtaining the sum over a of (a-1)^(-1) minus (a)^(-1), which is 1.

• Diego Viveros 10 months ago +2

By switching the order, we can find the value of the B sum, and then the A sum, and lastly we integrate.
The result of the B sum is just e, but started from 1 instead of 0, so it is e^x - 1.
Then we take the e^x -1 out of the A sum, as it does not depend on A, and the sum is a geometric sum, starting from 2 instead of 0, so we subtract the first 2 terms, leaving us with e^-x/(e^x-1)
Lastly, the integral is (e^x-1)e^-x/(e^x-1)dx, which is only e^-x from 0 to infinity, which is just 1.

• Phoenix Fire 10 months ago

1
Take the b! out of the integral. The integral then evaluates by parts to a^-(b+1)*b!. The b!'s cancel. The first geometric sum then evaluates to 1/(a^2-a). Fixing the second sum to start at 1 and factoring, we have 1/(a(a+1)) which we can manipulate into 1/a-1/(a+1). All the terms cancel right next to each other except for the first and last. That becomes 1/1-1/(infinity).

• Phoenix Fire 10 months ago

Fun algebra: 1/(ab)=(a+1)/(ab)-1/b
1/(ab)=1/(ab)+1/b-1/b=(b+ab)/(ab^2)-1/b=(1+a)/(ab)-1/b
Similarly: 1/(ab)=(1-a)/(ab)+1/b

• GreenMeansGO 10 months ago +1

With integration by parts, the all of the parts go to zero except for the final term -1/(a^(b+1)*e^(ax)) which goes to 1/a^(b+1). The sum over b is a geometric series which results in 1/a(a-1) which equals
1/(a-1)-1/a. The sum over a is a teliscoping series which equals 1.

• Allan Lago 10 months ago +2

I first switched the order between the sigmas and the integral and factored all of the factors not involving b:
int from 0 to inf (sum from a=2 to inf (e^(-ax) * sum from b=1 to inf x^b /b!))
Since e^x = sum from n = 0 to inf x^n /n!:
sum from b=1 to inf x^b /b! = e^x -1
By factoring e^x -1 out of the remaining sigma, the expression simplified to:
int from 0 to inf ((e^x-1) * sum from a=2 to inf 1 /e^(ax))
The inner sum is then sum of geometric series so we can use the formula: a_0 / (1-r) = (e^(-2x))/(1-e^(-x)) = (e^(-x))/(e^x -1)
int from 0 to inf (e^x-1) * (e^(-x))/(e^x -1) = int from 0 to inf e^(-x) = 1.

• Gabriel N. 10 months ago

First, let u=ax, to get the same sum of a^{-(b+1)}, since the integral becomes the gamma function, which cancels with the factorial. the b sum is a geometric series with ratio 1/a and initial value 1/a^2 which evaluates to 1/a(a-1). Now we need to sum 1/a(a-1) from 2 to infinity. We use 1/a(a-1) = 1/(a-1) -1/a to use à telescoping sum to cancel all but the 1.

• Hung Hin Sun 10 months ago +4

First we use integration by parts b times , we get integral of e^(-ax)/a^b, then after integrating it we obtain the result of 1/a^(b+1).
Then we consider the summation over b, it is a G. P. and its sum is 1/[a(a-1)] which can be rewrite as 1/(a-1) - 1/a. Then we consider the summation over a. From a = 2 to a = N, the result is 1 - 1/N, as N tends to infinity, we get the result to be 1.

• Benjamin Wang 10 months ago

Looks like we did it the hard way? Lol. Everyone else switched the sum

• Benjamin Wang 10 months ago

It evaluates to UNITY.
Denote I(a,b)=x^b.e^{-ax}. By parts we get I(a,b)=b/a.I(a,b-1). Together with I(a,0)=1/a we have I(a,b)=b!/a^(b+1). Thus we simplify to the sum from two to infinity of (1/a^2 +1/a^3 +...). By geometric series the bracket is 1/((a)(a-1)). By telescoping the sum is ONE.

• M. Shebl 10 months ago

Let's evaluate this!
This equal to the series (Switch series):
sum_{b=1}^{\infty} 1/b! sum_{a=2}^{\infty} L(x^b)
Where L(x^b) is the laplace transform at s=a.

sum_{b=1}^{\infty} 1/b! sum_{a=2}^{\infty} b! a^{-b-1}
= sum_{b=1}^{\infty} sum_{a=2}^{\infty} a^{-b-1}
Switch series again
sum_{a=2}^{\infty} 1/a sum_{b=1}^{\infty} (1/a)^b
Use sum_{b=1}^{\infty} (1/a)^b = 1/a/(1-1/a) = 1/(a-1)
sum_{a=2}^{\infty} 1/a(a-1)
This equals sum_{a=1}^{\infty} 1/a(a+1)
1/a(a+1)=1/a-1/(a+1)
sum_{a=1}^{\infty} 1/a(a+1) = 1/1-1/2+1/2-1/3+...=lim(1-1/n) as n goes to inf
=1

• SoW Lone Archer 10 months ago +8

The answer is you, baby. ❤

• Benjamin Wang 10 months ago

LetsSolveMathProblems Please cover all these relevant solution methods (or as many as you can) that viewers proposed. 1. Integration directly, 2. switching sums to simplify;3 Gamma .4 Laplace. Problems with more solutions are all the more fascinating and profound

• LetsSolveMathProblems  10 months ago +6

It is me, the one and only one. =)

• Problematic(puzzle channel) 10 months ago +1

first comment!

• Benjamin Wang 10 months ago

well coming to think about it the answer IS contained in the word '1st'...