# Challenge 58: A Chain of Logarithms

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• Published on Sep 13, 2018
• Congratulations to Gabriel N., GreenMeansGO, Benjamin Wang, pichutarius, Daniel Darroch, Cobalt314, Evyatar Baranga, Sourin Chatterjee, Woof, and Nicola C for successfully solving the last week's math challenge question! Gabriel N. was the first person to solve the question.
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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.

## Comments • 35

• GreenMeansGO Year ago

I finally got around to doing this problem. The trick is to simplify as much as possible. In the end, I found that since p_n = 2/n*p_(n-1) then the sum turns into half of p_2 times the sum from j=2 to infinity of 2^j/j!. P_2 equals 2 to half is 1. So we get 1 times the series which is the series for e^2 minus the terms for j=0 and j=1 so you get e^2-3.

• Benjamin Wang Year ago

e^2 -3
I’ve been away from home so I’m super late on this one. The product from 1 to n-1 of a sub k cancels down to (ln(n+1)/ln2)^2 times 2^{n-1}/{(k-1}!. Thus P sub n simplified to be similar to the definition of e^x, which gives the answer.

• Random Person Year ago

We first simplify all the logs with change of base and get that
a_k=(2 log^2(k+2))/(k log^2(k+1))
P_n=(2 log^2(2))/(n log^2(n+1))*\prod_{k=1}^{n-1} a_{k}
(\prod_{k=1}^{n-1} a_{k} is the product from k=1 to n-1 of a_k)
By deducing an expression for the product part of P_n, we get that
\prod_{k=1}^{n-1} a_{k}=(2^(n-1)log^2(n+1))/((n-1)!*log^2(2))
Multiplying this with the first part of P_n yields
P_n=2^n/n!
And summing this from n=2 to infinity yields
e^2-3.

• Please solve or discuss standard question 58 from NS7UC

• WaltherSolis Year ago

e^2-3, p_n can be simplified to 2^n/n!, you just need to apply a lot of logarithms properties

• e^2 -3
becouse a,k can be written as (log,(k+1) (k+2))/(log,(k+2) (k+1)) and log,a b / log,b a = (log,a b)^2, so a,k=((log,(k+1) (k+2))^2.
The product may be written as (log,2 (n+1))^2 , so P,n=log,(n+1) 2 * log,2 (n+1) {that is equal to 1} times 2^n /n!
The summation is a Taylor series that is equal to e^2 -3

• e^2 -3
Aryan Jain: "Prove it"

• e^2 - 3, only applying logarithm's properties. Results the summation of n = 2 to infinity, and that is the summation of n = 0 to infinity minus the terms of n = 0 and n = 1, which are 1 and 2 respectivly. We know that the summation of n = 0 to infinity of (2)^n/n! is e^2, whose difference with 3 is equal to e^2 - 3!

• About Math Year ago

e^2-3

• Aryan Jain Year ago

Prove it

• Lasitha Nirmitha Year ago

first consider the general term of a, we can simplify it to (logk+1 (k+2))\(logk+2 (k+1)) ×(2/k) by using logarithm rules. using that and using the formula loga(b)=logc(b)/logc(a) we can find pn to be 2^n/n!. by summing pn from 2 to infinity we get the answer e^2-3

• Using simple properties of log the expression simplifies to (ln2)(2^n/nlog(n+1)) then apply ratio test which gives 2 and hence it diverges.

• Angelo D. Year ago

Beautiful problem! I got the same answer e^2-3, but I'm late... Thanks for your challenging questions and congratulations for your great work!

• Aryan Jain Year ago

Prove it

• Nicola C Year ago

The answer is e²-3.
Pₙ simplifies with a lot of log properties and all logs cancel out, leaving 2ⁿ/n!.
We have to evaluate ∑ from 2 to ∞ of 2ⁿ/n!.
If we consider the sum from 0 to ∞, it is the power series of e².
So we have to subtract 2⁰/0! and 2¹/1!, which is just 3.
Therefore our final answer is e²-3 and we are done.

• Kartik Sharma Year ago

e^2-3

• Aryan Jain Year ago

Prove it

• Adwait Kulkarni Year ago

The answer is (e²)-3
First after using the change of base property for logarithms we change a sub k to [2*(ln(k+2))²]/[n*(ln(k+1))²]. We observe that the logarithmic term in the numerator of a of k is the same and the denominator of a of (k+1). Therefore while evaluating the product after many cancellations we the get product to be equal to [(2^(n-1))*(ln(n+1))²]/[(n-1)!* (ln2)²]
Similarly manipulating the the terms before the Product in P of n we get that Pn equals [2^n]/n!
The sum is similar to the series expansion of e², so after adding and subtracting [(2^0)/0! + (2^1)/1!] We get the answer of the summation as e² - 3

• Rajat Khandelwal Year ago

I think ans is e^2 - 3
Because when we solve for ak we get (2/k)(log base(k+1) k+2)^2 and when
We product it from k=1 to n-1 weget expression 2^n-1/(n-1)!(log 2(base) n+1)
Now when we solve for pn we get 2/n(log (n+1)(base) 2) and when we product it with ak we get 2^n/n! Now summation of this is clearly a divergence sequence and it get us ans of e^2-3

• Aswini Banerjee Year ago

(log(x^n) to the base of y^m) means n/m*logx to the base of y. By this rule we get a_k=(2/k)*(log(k+2)/log(k+1))^2
Now the product of a_k is telescopic and equals to (2^(n-1)/(n-1)!)(log(n+1)/log2)^2
Now the first term in p_n equals to (2/n)*(log2/log(n+1))^2
So p_n =2^n/n!
Now n=2 to inf sum of p_n is equal to (sum of (2^n/n!) from n=0 to inf)-2^0/0!-2^1/1!=e^2-3

• Hiren Bavaskar Year ago

Answer is e^2 -3
Using properties of log, ak simplifies to
2(log(k+2))^2/k(log(k+1))^2
Finally using properties of log and simplifying Pn, we get Pn as
2^n/n! Hence the series can summed using the expansion of e and noticing the value of e^2= 1+ 2/1+ 2^2/2!....
Hence our series which is 2^2/2! + 2^3/3!..... comes to e^2-3

• Albanovaphi7 Year ago

En efecto , es fácil comprobar usando propiedades de logaritmos que a_k = (2/k)(log(k+1)(k+2))^2 (donde log(a)(b) significa que a es la base y b la potencia), y el productorio PI(desde k=1)(hasta n-1) = [(2^(n-1))/(n-1)]*(log(2)(3)*log(3)(4)...log(n)(n+1))^2 = [(2^(n-1))/(n-1)]*(log(2)(n+1))^2. Reemplazando esto en p_n los logaritmos se cancelan quedando que p_n = (2^n)/n! y está claro que la sumatoria (que llamaremos S) se obtiene de la expresión e^x=(x^0)/0! + (x^1)/1! + ... + (x^n)/n! donde x=2. Entonces e^2 = (2^0)/0! + (2^1)/1! + (2^2)/2! ... + (2^n)/n! = 1 + 2 + ((2^2)/2! ... + (2^n)/n! ) = 3 + (S), luego S converge a (e^2 - 3), es decir; S=(e^2 - 3).

• Woof Year ago

Answer is : e^2 - 3

a_k is simplified to (2/k)(log_k+1 k+2)^2
Πa_k = (2^(n-1)/(n-1)!)(log_2 n+1)^2
P_n = 2^n/n!
ΣP_n is very similar to the taylor series of e^x at x=2, but subtracting the first 2 term
therefore, ΣP_n = e^2 -1 -2 = e^2 -3

• Nice problem. I think the answer is e^2-3

• Aryan Jain Year ago

Prove it.

• Since log_a (b^c) = c*log_a (b) and log_a (b) = log b / log a, after simplifying some terms, we can conclude that a_k = 2/k * (log (k+2) / log (k+1))^2. The product of all the a_k is then equal to 2^(n-1)/(n-1)! * (log (n+1) / log 2)^2. Simplifying in a similar way the first fraction, we obtain 2/n * (log 2 / log (n+1))^2, so all the logarithms disappear and we get P_n = 2^n/n!
On the other hand, we know that the sum (from n = 0 to infinity) of x^n/n! is equal to e^x, so the sum of P_n from n = 2 to infinity is equal to e^2-2^0/0!-2^1/1! = e^2 - 3.

• Bin Vissotsak Year ago

Ans: e^2

• Allan Lago Year ago

I got e^2 -3.
a_k = (2k/(2k-1) log_(k+1) (k+2))/(k^2 /(2k-1) log_(k+2) (k+1)) = 2/k *(log_(k+1) (k+2))^2.
P_n = 2/n *(log_(n+1) (2))^2 * product{a_k}
Using the property log_a (b) * log_c (d) = log_a (d) * log_c (b):
P_n = 2/n * product{2/k}
ln(P_n) = n * ln(2) - sum {ln(k)} -ln(n)
ln(P_n) = n*ln(2)-ln(n!)
P_n = 2^n /(n!)
sum{P_n} from n = 2 to inf = sum{2^n /(n!)} from n = 2 to inf = e^2 -3.

• Beshoy Nabil Year ago

Ans = (e^2)-3 approximately 4.389056
Since log (a^b) to base (c^d) = (b/d)* log (a) to base (c) & log (a) to base (c)= log (a)/log(c)
Then a_k = (2*((log(k+2))^2))/(k*((log(k+1))^2))
So the product of these terms from k=1 to n-1 will telescope giving ((2^(n-1))*((log(n+1))^2))/(((n-1)!)*((log(2))^2)) then multiplying it by the rest of P_n expression after applying the first rules will end
at P_n = 2^n / n!
Since e^x = Summation of x^n/n! from n=0 to infinity
Then e^2 = 1+2/1! + Summation of 2^n/n! from n=2 to infinity
Then Summation of 2^n/n! from n=2 to infinity = (e^2)-3

• Gabriel N. Year ago +2

Beautiful problem. I believe it's e^2-3, but I'll leave this one alone.

• Gabriel N. Year ago +2

... I said I didn't want to participate this week.

• Essentials Of Math Year ago +10

e^2 - 3.
Using some log properties, a_k simplifies to 2/k * ln^2(k+2)/ln^2(k+1). Similarly, the expression outside of the product simplifies to 2/n * ln^2(2)/ln^2(n+1). It can then be observed that P_n telescopes down to 2^n / n!. We know e^x = 1 + x + x^2/2! + x^3/3! +..., So e^2 = 1 + 2 + (sum of P_n), so rearranging gives a result of e^2 -3.

• Yugesh Keluskar Year ago

You got a new subscriber

• • 