# 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.

GreenMeansGO9 months agoI 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 Wang9 months agoe^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 Person9 months agoAnswer is e^2-3.

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.

Siddhartha Shree kaushik9 months agoPlease solve or discuss standard question 58 from NS7UC

Please Google NS7UC

WaltherSolis9 months agoe^2-3, p_n can be simplified to 2^n/n!, you just need to apply a lot of logarithms properties

Antonio Raddakovic9 months agoe^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

wherestheshroomsyo9 months agoe^2 -3

Aryan Jain: "Prove it"

Mathias ICARTE MANCILLA9 months agoe^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 Math9 months agoe^2-3

Aryan Jain9 months agoProve it

Lasitha Nirmitha9 months agoanswer is e^2-3

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

Rudra Pratap Singh9 months agoUsing 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.9 months agoBeautiful 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 Jain9 months agoProve it

Nicola C9 months agoThe 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 Sharma9 months agoe^2-3

Aryan Jain9 months agoProve it

Adwait Kulkarni9 months agoThe 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 Khandelwal9 months agoI 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 Banerjee9 months 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 Bavaskar9 months agoAnswer 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

Albanovaphi79 months agoEn 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).

Woof9 months agoAnswer 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

Quahntasy - Animating Universe9 months ago^{+2}Nice problem. I think the answer is e^2-3

Aryan Jain9 months agoProve it.

Abraham del Valle Rodríguez9 months agoSince 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 Vissotsak9 months agoAns: e^2

Allan Lago9 months agoI 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 Nabil9 months agoAns = (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.9 months ago^{+2}Beautiful problem. I believe it's e^2-3, but I'll leave this one alone.

Gabriel N.9 months ago^{+2}... I said I didn't want to participate this week.

Essentials Of Math9 months 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 Keluskar9 months agoYou got a new subscriber

adandap9 months agoEssentials Of Math - ah, thank you. Expressing all of the logs in terms of base 2 was the missing link for me. (BTW, I think your answer here is missing that it's the *square* of log_2 (n+1) that appears in the expression for the coefficient and the a_k)

Rafael Assunção9 months ago^{+1}Primerão dos BR HUEHUEHUEHUEHUEHUE