# Challenge 86: Can You Sum the Reciprocals of (6n)!! ?

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**Published on Mar 27, 2019**- Congratulations to quak quak, Albert Estellé Caballer, Rohit Agarwal, Minh Cong Nguyen, Intergalactic Bajrang dal, and mstmar for successfully solving the last week's math challenge question! quak quak was the first person to solve the question.

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A Name2 months agoTheta = sqrt(3)/4

Note that the summation of 1/(2n)!! from n=0 to infinity is 1 + 1/2 + 1/2*4 + 1/2*4*6 + ...

Also note that this is the same sum as you obtain from plugging in x=1/2 to the taylor series expansion for e^x (and thus the sum is equal to sqrt(e))

This is a tangent, but it was an essential step for me in getting to the final answer. Consider what the infinite sum of 1/(4n)!! is. Note that plugging in x=-1/2 to the taylor series expansion for e^x to yield 1 - 1/2 + 1/2*4 - 1/2*4*6 +...

Taking e^1/2 + e^(-1/2) will give 2*(1 + 1/2*4 + 1/2*4*6*8 + ...) and the infinite sum can then be easily found.

Using the same strategy as above it would be nice if we could create an infinite sum that looks like 1 - 1/2 - 1/2*4 + 1/2*4*6 - 1/2*4*6*8 - ...

Since the +/- signs have a repeating cycle of 3, the 3rd root of unity seems like its worth a shot. Using w = -0.5 + i*sqrt(3)/2, e^(1/2*w) = 1 + w/2 + w^2/2*4 + 1/2*4*6 + w/2*4*6*8 +....

This is kind of like what we want, but we have w and w^2 instead of -1's. But since w^2 + w = -1, we can also look at e^(1/2*w^2) = 1 + w^2/2 + w/2*4 + 1/2*4*6 + w^2/2*4*6*8 + ....

Adding the two we obtain the desired infinite sum plus 1 + 1/2*4*6 + 1/2*4*6*8*10*12 which is in fact the infinite sum of 1/(6n)!!. Adding this to sqrt(e) = 1 + 1/2 + 1/2*4 + 1/2*4*6 +... we end up with:

sqrt(e) + e^(1/2*w) + e^(1/2*w^2) = 3*(1 + 1/2*4*6 + 1/2*4*6*8*10*12 +...)

Some algebra combined with e^ix = cos(x) + i*sin(x) we find that the infinite sum is equal to sqrt(e)/3 + 2*cos(sqrt(3)/4)/3*e^1/4, or theta = sqrt(3)/4

Mohammad Yusuf2 months ago^{+1}Is the answer pie by 2

kinshuk dua2 months agoWhat I did was work out the first 3 terms of the series, it can be seen that since the number in the denominator are all even number we can take 2 common 3*n times because of which the series can be written as sigma ((2)^-3n)/(3n)! which is super similar to Taylor series expansion for e^x now we can take the n out making the sum sigma ((1/8)^n)/(3n!)) which is obviously smaller than e^(1/8). This is where I'm stuck, I cannot find out a way to take out the 3 as I don't think there is a property of factorial that allows me to take out the 3. I'm hoping the sum can be somehow converted to an exponential function and then we can solve the rhs to for e^something and equalize both side and get the value of cos theta and then theta

梁瀚文2 months agon= 0 1/(6+n)!! = 1

n= 1 1/(6+n)!! = 1/(6*4*2) ===> f(x) = 1

n= 2 1/(6+n)!! = 1/(12*10*8*6*4*2)

e= 2.71 e^0.5 = 1.65... e^0.25=1.28....

1 = 1.65/3 + [2/(3*1.28)]*cos

cos = 0.865

answer = 30 o

Sachin Sahay2 months agotheta = sqrt(3)/4.

Let z = (e^(2*pi*i / 3))/2 and z' = (e^(2*pi*i / 3))/2. Plug z and z' into the taylor series for e^x, add them and rearrange to get e^z + e^z' = 3(sum of 1/(6n)!!) - sqrt(e), so the sum of 1/(6n)!! is (e^z+e^z'+e^0.5)/3. Substitute the original definitions of z and z' into this and simplify with De Moivres and separating the exponents into real and imaginary parts. The imaginary terms cancel out, leaving the given sum as sqrt(e)/3 + 2 cos (sqrt(3)/4) / 3(e^0.25), so theta = sqrt(3)/4.

Avirup Dey2 months agoUse taylor expansion of e^(0.5x) and put 1,w,w^2...Then add all three series. ..Ans will be [e^(0.5) + {2*e^(-0.25)*cos(✓3/4)}]/3

Haradhan Datta2 months agoAnswer=√3/4.To solve the problem, the following informations & formulae are to be needed:(1) Concept of double factorial. (2) Exponential series: (e^x=1+X/1!+x^2/2!+x^3/3!+........upto infinity.(3) Euler's formula:e^x=1+x/1!+x^2/2!+x^3/3!+.......up to infinity.(4) Properties of cube roots of unity :w^3=1,1+w+w^2=0,w^3k=1,w^3k+1=w,w^3k+2=w^2 for all k belongs to integers and w=(_1+i√3)/2, w^2=(_1_i√3)/2. Now (6n)!!=(6n)(6n_2)(6n_4)......6.4.2=(2^3n).(3n)(3n_1)(3n_2).......3.2.1=(2^3n).(3n)!.Let the sum be S. Therefore,S=summation from 0 to infinity 1/(6n)!!=summation from 0 to infinity 1/((2^3n)(3n)!)=summation (1/2)^3n/(3n)!=summation x^(3n)/(3n)! , (taking X=1/2) =1+x^3/3!+x^6/6!+x^9/9!+.........up to infinity. Now, e^x=1+X/1!+x^2/2!+x^3/3!+x^4/4!+x^5/5!+x^6/6!+..........up to infinity..........(I) Replacing X by wx &(w^2)X, we have, e^wx=1+(wx)/1!+(w^2.x^2)/2!+(w^3.x^3)/3!+(w^4.x^4)/4!+(w^5.x^5)/5!+(w^6.x^6)/6!+.......... up to infinity. ........(ii) and e^(w^2.x)=1+(w^2.x)/1!+(w^4.x^2)/2!+(w^6.x^3)/3!+(w^8.x^4)/4!+(w^10.x^5)/5!+(w^12.x^6)/6!+........ up to infinity. .......(iii). Adding (I),(ii),(iii) and using the properties of cube roots of unity,we get , e^x+e^(wx)+e^(w^2.x)=3(1+x^3/3!+x^6/6!+........up to infinity)=3S. So, S=(e^x+e^(wx)+e^(w^2.x))/3=(e^(1/2)+e^(w/2)+e^(w^2/2))/3 =√e/3+(1/3)(e^((_1+i√3)/4)+e^((_1_i√3)/4)=√e/3+(1/3)((e^(_1/4)(e^(i√3/4)+e^(_i√3/4))= √e/3+(1/3e^(1/4))( cos(√3/4)) +i sin(√3/5)+cos(√3/4)_ I sin(√3/4))=√e/3+(2/3e^(1/4)) cos(√3/4) . But it is given that,S=√e/3+(2/3e^(1/4)) cos(theta). Hence, comparing we have, theta=√3/4.

Haradhan Datta2 months agoTwo days ago,I solved the problem. But I have been mistaken (3) Euler's formula: e^x=cos(X)+i sin(X), I=√(_1).I'm very sorry for the mistake. Thanks.

Dmitry Miserev2 months agotheta = 3^{1/2}/4

(6n)!! = (2^{3 n})* (3n)!. Consider function:

f(x) = sum_0^\infty x^{3 n}/(3 n)!

It satisfies the differential equation f'''(x) = f(x).

Substituting f(x) = e^{s x}, one gets s^3 = 1, so

s1 = 1, s2 = e^{2 pi i/3}= -1/2 +i sqrt(3)/2, s3 = e^{-2 pi i/3} = -1/2 - i sqrt(3)/2.

As we search for real-valued function, we can write

f(x) = a e^x + b e^{-x/2} cos(x*sqrt(3)/2).

As f(0) = 1, f'(0) = f''(0) = 0, we get a = 1/3, b = 2/3.

So, f(x) = (e^x)/3 + (2/3) e^{-x/2} cos(x*sqrt(3)/2).

The sum is equal to f(1/2), so theta = sqrt(3)/4.

Mushishi28722 months agotheta = sqrt(3)/4

Let S = Sum of 1/(6n)!! from n = 0 to infinity

(6n)!! = 2*4*6*...*(6n-2)*(6n) = (3n)!*2^(3n)

So, S = 1 + 1/(3!*2^3) + 1/(6!*2^6) + ...

Define function: y(x) = 1 + x^3/3! + x^6/6! + ...

then, y(1/2) = S

In order to find y(x) we notice that:

y'(x) = x^2/2! + x^5/5! + x^8/8! + ...

y''(x) = x + x^4/4! + x^7/7! + ...

Therefore: y + y' + y'' = e^x

By solving the 2nd order ODE , knowing that y(0) = 1 and y'(0) = 0, we get:

y(x) = (e^x)/3 + (2/(3e^(x/2)))cos((sqrt(3)/2)x)

y(1/2) = (e^(1/2))/3 + (2/(3e^(1/4)))cos(sqrt(3)/4)

Thus, theta = sqrt(3)/4

Proof by Meme2 months agoWow, what an excellent problem! The complex route of substituting the Taylor for e^[x/2] with e^[(e^2πi/3)(x/2)] means that the chain rule will determine only multiples of 3 to have real coefficients.

Dealing with all the imaginary parts is completely beautiful because 2π/3 and 4π/3 are coincidentally the only possible angles for the other roots, and they happen to be conjugates of each other!

I suspect that we can get other nice results from ∑1/(2kn)!! as long as k allows for symmetry with the roots of unity (easy conjugates).

Or, it may be that k=3 is the only “nice” solution because all other regular polygons besides the triangle will not have the same real symmetry that allows extensive factoring of the Taylor’s degrees that aren’t a multiple of 3.

This was just an epic problem. :D

Serengeti Ghasa2 months agothe answer is π/2

if we take the value of the sum as y and x=1/2

then y+dy/dx+d²y/dx²=eˣ=√e

the equation hold good if y is √e/3

so cosθ is 0 implias θ is π/2

santosh tripathy2 months ago^{+4}So the Answer is root(3)/4.

first of all note that the summation can be written as Sum of 1/2^(3r) into 1/(3r)! where n! represents the product of first n natural nos.

since e^x=1+x+x^2/2!+x^3/3!... putting x=1,w,w^2 (where w is a complex root of unity) and adding we get 3 times the summation required= e^0.5+e^0.5w+e^0.5(w^2)

now arranging this expression as required we see that cos(theta)=( e^((2w+1)/4)+e^((2w^2+1)/4))/2 since 1+w+w^2=0.Thus w=(-1+i sqrt(3))/2 putting this in the expression and using the eulers identity that e^iz=cosz+isinz follows that cos(theta)=cos(root3/4)..since theta lies in b/w 0 and pie, THUS THETA= root3/4

8 Bit Thoughts2 months ago7. The answer is 7 right?

Fun With Vaishnavi2 months ago^{+1}Theta=(3^1/2)÷4

Taking some hint from EOM ,

I solved it like him but try a lot before seeing his comment but after a small hint I finished up by own .As no explanation because EOM explained it!!!

Hats off to the EOM.👍👍👍

Calcul8er2 months agoBasic outline of a proof:

Define f(x) = sum from 0 to ∞ of x^(3n)/(3n)! then the answer is f(1/2) which can be seen by expressing (6n)!! as 2^(3n) (3n)!. Differentiating term wise you can find that f”(x)+f’(x)+f(x)=e^x. Solving for the general solution then using the conditions f(0)=1 and f’(0)=0 gives an explicit form for f(x). Plugging in x=1/2 gives the desired solution

jcfgykjtdk2 months agoPi/2

Chan Dan2 months agoIs higher mathematics allowed or i have to solve using highschool mathematics, plz reply anyone ?

Benjamin Wang2 months ago^{+1}Niranjan kumar I don’t think there’s a restriction

aby p2 months ago^{+1}i split the summation like we do in patial fractions in integration then we get one summation as (1/2)^n/n! the whole divided by 3 from 0 to ifinity which is (e)^1/2/3 and then the other summation is 2/3(2n-1)(2n-3)... and then on the RHS we use the cosx expansion and then we get the ans as (3)^1/2/4 we can also try one more method when we open the summation and put values of n we get the summation as 1/(3n)! x (1/2)^3n then on the RHS we use the expansion of cosx and and we get the ans as root3/4

Paco Libre2 months ago(2n)!! = 2^n*(n!) (extract a 2 from each term) so the sum can be written as sum of 1/(2^(3n)*(3n)!). This sum is a special case of the maclauran series sum of x^(3n)/(3n!) where x=1/2. Differentiating three times gives us sum from 3 to infinity of (3n)*(3n-1)*(3n-2)*x^(3n-3)/(3n)! = x^(3n-3)/(3n-3)! which gives back the original sum after shifting the index. This means that the series solves the differential equation y'''=y. The roots of the characteristic equation are the cube roots of unity giving a solution of y=C1e^x+C2e^(-x/2)*cos(root3/2*x)+C3e^(-x/2)*sin(root3/2x). For the constants, look to the values of the power series to get y0=1, y'0=0, and y''0=0. Plugging in these values gives y=(e^x+2e^(-x/2)*cos(root3/2*x)/3. Finally, plug in x=1/2 for the original sum to get root2(e)+2/root4(e)*cos(root2(3)/4). Thus, theta=sqrt(3)/4.

aby p2 months agoFirst we open the (6n)!! and write it as (6n-2)(6n-4)(6n-6).... then we take 2 common and write it in the numerator as (1/2)^n and then we take (n-1)(n-2)(n-3)... on one side of the denominator and (2n-1)(2n-3)... and then we split the summation into two parts and therefore we get the ans as (3)^1/2/4

LetsSolveMathProblems2 months agoAs of now, your solution is not quite comprehensive enough to be considered for recognition. Could you elaborate more on how exactly you split "the summation into two parts" *and* how it led to the final answer?

* Please do not edit your comment, but instead post a new comment with the modified solution.

adandap2 months ago^{+2}I tried this for a couple of hours but didn't get the insight that Essentials of Maths showed below. Here's how I failed (now there's a title for my autobiography!).

The first thing to do is to turn the !! into something more familiar, and it's not hard to see that (6n)!! = 2^(3n) (3n)! So the series we want is Sum( 1/(2^(3n) * (3n)!) That's suggestive of an exponential series with x=1/2, with two out of every three terms missing. I tried writing series expansions for e^(1/2) and e^(-1/4) and then trying to find a value for cos(theta) so that all of the unwanted terms vanished. Let me save everyone else some pain: don't do it that way! As EoM points out, you can generate the series you want by a clever bit of complex analysis.

Oh well, onwards and upwards for next week. This was a nice problem, and at least a good workout for my math brain - which is sadly like the 'before' guy in the Charles Atlas ads of old.

adandap2 months ago@Peter Yes, that was a transcription error. I actually calculated it properly. But thanks for the heads up. I'll edit it.

Peter2 months agoYou made a mistake in that (6n)!!=2^(3n)*(3n)!, not 2^n*(3n)!.

LetsSolveMathProblems2 months ago^{+3}I personally think your approach is interesting, as well, even if it did not quite work in the end. (I would love to read your autobiography, if you happen to publish it.) Besides, often in mathematics, knowing why something does not work sheds more light on the problem than knowing the correct solution. I believe Wiles realized how to fix the mistake in his proof of Fermat's Last Theorem the moment he understood why exactly in his original approach was not good enough.

For this particular problem, I recommend looking up "Roots of Unity Filter," which is a nice technique that can be used to sum up the coefficients of a polynomial (mod n). =)

Ben Burdick2 months agoWell, I know I'm wrong, but I'm going to post my work here because I did my best and I want to know where I went wrong.

Start by seeing that (6n)!! = 2^(3n)*(3n)!, by realizing that 6n!! can be written as (2*3*n)!!. This becomes 2(3n)(2(3n)-2)(2(3n)-4)...(2). Factoring out a 2 shows that 2^(3n)*(3n)! is indeed correct.

Compare this to the taylor expansion for e^x. By plugging in 1/2 for x, we get close to the right answer with the infinite sum of (1/2)^n/n!. By using the substitution 3k=n, we get the formula 2^(-3k)/(3k!). The bounds on our sum remain the same, and this does not change the value. ***

This shows that our sum is equal to e^1/2, and by doing algebra we can see that our answer should be arccos(e^3/4), which is undefined.

I think my mistake happened with my substitution. Any feedback is appreciated.

adandap2 months agoHi Ben. As you guessed, you can't change the summation variable like that. Try going back to the series with terms x^(3n) /(3n)! and x^n/n! and writing out the first four or five terms of the series before and after that change and you'll see why - you have added all of the 'missing' powers of x back in.

Hiren Bavaskar2 months ago√3/4 is the answer !

Let the sum be denoted by S.

By writing the terms and taking 2 out of each even digit we get S= summation of 1/[(2^(3r)) * (3r)!] from 0 to infinity.

We have to skip 2 terms in expansion of e^ (1/2).

So put x=1, w/2, (w^2)/2 {where w is cube root of unity} in expansion of e^x and add. This will be 3S by the addition. So our answer is (1/3) *[e^0.5 + e^0.5w + e^0.5( w^2) ] . Using e^i theta expansion and simplifying we get theta = √3/4

Hiren Bavaskar2 months agoOne more correction: We have to skip two terms one after the another in expansion of e^0.5 [ eg: we don't want 1/2*(1!) and 1/2^2*(2!) and 1/2^4*(4!) and 1/2^5*(5!) so on.. ]

Hiren Bavaskar2 months agoCorrection: Put x = 0.5, w/2 and w^2/2 in e^x. I wrote 1 in a hurry😭

fmakofmako2 months agoLet S(x) = Sum from n=0 to inf of x^(3n)/(3n)!

Notice S + S' + S'' = e^x by Taylor expansion

The complementary solutions of S are e^rx where r is the solution of 1+r+r^2=0, which is on the unit circle r = e^2pi i/3,e^-2pi i/3.

The particular solution is found by undetermined coefficients. It is e^x/3.

S(x) = e^x/3 + c1 e^ r1x + c2 e^r2x = e^x/3 + e^(-x/2) [(c1 + c2) cos (sqrt(3)/2x) + (c1-c2) i sin (sqrt(3)/2x)]

When x=0, S(x)=1 from the series definition so c1 + c2 = 1. Also the definition of S for real x is real, so c1 - c2 = 0.

S(x) = e^x/3 + e^(-x/2)cos(sqrt(3)/2x)

S(1/2) is our series so sqrt(3)/4 is theta.

Peter2 months agoThe answer is theta = sqrt(3)/4. Firstly I wrote the original summation as the sum of x^(3n)/(3n)! where x=1/2. Then if we let y'' = sum from 0 to inf of x^(3n)/(3n)! we can write the differential equation:

y''=e^x-y'-y

Which can be solved to give

y''=e^x/3 + (2cos((sqrt(3)/2)x))/(3e^(0.5x))

By using the boundary conditions y(0)=0, y'(0)=0 etc. In the case where x=1/2 the sum is e^0.5/3 + (2cos(sqrt(3)/4))/(3e^0.25). So theta = sqrt(3)/4.

Essentials Of Math2 months ago^{+14}Theta = sqrt(3)/4.

First, it can be shown that (6n)!! = 8^n * (3n)!, simply by factoring out a two from every term in the product. (Special care is taken when n = 0, but of course (0)!! = 1 = 8^0 * (0!).)

Then the desired sum is Sum from n = 0 to infinity of 1/(2^(3n) * (3n)!).

Recall the Taylor series of e^x = Sum from n = 0 to infinity of x^n / (n)!. What we need is a way to extract only multiples of three from the expansion of e^(1/2).

Let w = e^(2pi i /3) = -1/2 + sqrt(3)/2. Then w^2 = -1/2 - sqrt(3)/2, 1 + w + w^2 = 0, and w^3 = 1. These properties induce the so-called roots of unity filter which we need.

Our desired sum is thus equal to 1/3 (e^(1/2) + e^(w/2) + e^(w^2/2)). Plugging in w and w^2, we get

1/3(e^(1/2) + e^(-1/4) [e^(i sqrt(3)/4) + e^(-i sqrt(3)/4)]). It is well known that e^(ix) + e^(-ix) = 2cosx, and we see that our sum has the form

e^(1/2) /3 + 2/(3e^(1/4)) * cos(sqrt(3)/4)). Thus the theta we are looking for is sqrt(3)/4.

Kien P.S.2 months ago@Essentials Of Math Oh! I got it! Thanks!

Essentials Of Math2 months ago^{+1}@Kien P.S. no, because there are (3n) terms in the product of (6n)!!, Each of which is even, so you can factor a 2. 3n factors of 2 makes 2^(3n) or 8^n

Kien P.S.2 months ago@Essentials Of Math I thought (6n)!! = 2 * (3n)!

Gabriel N.2 months agoAnswer: sqrt(3)/4

Let G(x)=e^x

Since (2n)!!=2^n * n!

We get that the sum is the exponential series Σx^k/k!, with only the terms where k is a multiple of 3 and with x= 1/2. By the section formula, it is equal to 1/3 * (G(1/2) + G(1/2 * w) + G(1/2 * w^2)), where w is the third root of unity. Using Euler’s Formula, we get the result.

Avinash Kumar2 months ago^{+2}1st view