# How to calculate the following limit? [closed]

Calculate the following limit where $n \in \mathbb{Z}$ and log is to the base $e$ $$\lim_{x\to\infty} \log \prod_{n=2}^{x} \Bigg(1+\frac{1}{n}\Bigg)^{1/n}$$

## closed as off-topic by user99914, Travis, user147263, Claude Leibovici, user223391 Jul 11 '15 at 6:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Travis, Claude Leibovici, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

• Please, write us your thoughts about the problem and what you have tried. – Blex Jul 4 '15 at 8:14
• Notice, that logarithm is a continuous function. – luka5z Jul 4 '15 at 8:49
• Yes i know that limit can go inside beacause of continuity of log... – user252347 Jul 4 '15 at 8:52
• It can "go inside" only if the limit $\lim_{x\to\infty}\prod_{n=2}^{x} \Bigg(1+\frac{1}{n}\Bigg)^{1/n}$ exists (if finite). – luka5z Jul 4 '15 at 8:53
• also it is enough to calculate this product 2^(1/2).3^(1/6).4^(1/12).5^(1/20)............... – user252347 Jul 4 '15 at 8:53

Hint:

Expressing $\ln\left(1+\dfrac1n\right)$ by Taylor expansion leads to

$$\sum_{n=2}^\infty\frac1n\left(\frac1{n}-\frac1{2n^2}+\frac1{3n^3}\cdots\right)=\sum_{n=2}^\infty \frac1{n^2}-\frac1{2n^3}+\frac1{3n^4}\cdots=\sum_{k=2}^\infty\frac{(-1)^k\zeta(k)}{k-1}.$$

Not really easier.

• interesting.....Thanks – user252347 Jul 4 '15 at 9:05
• As no closed form of the summation of $1/n^3$ is know, you can't really find a exact answer. – Ahmed S. Attaalla Jul 4 '15 at 9:11
• @AhmedS.Attaalla: there is indeed no closed form for $\zeta(3)$, but that doesn't mean that there is no closed form for a sum of $\zeta$'s. For instance, $\sum(\zeta(n)-1)=1$ and $\sum(\zeta(n)-1)/n=1-\gamma$. – Yves Daoust Jul 4 '15 at 9:14
• As we know the sum of $1/n^2$ is $pi^2/6$ if we're able to come up with a closed form of this summation than we would be able to come up with a closed form for $1/n^3$. – Ahmed S. Attaalla Jul 4 '15 at 9:16
• Or am I wrong @Yves Daoust – Ahmed S. Attaalla Jul 4 '15 at 9:19

For the confirm that there is no closed form for series found by Yves Daoust, we can use the identity $$\sum_{k\geq2}\frac{\left(-1\right)^{k}x^{k}\zeta\left(k\right)}{k}=\gamma x+\log\left(\Gamma\left(x+1\right)\right),\,\,-1<x\leq1.$$ Take the derivate to get $$\sum_{k\geq2}\left(-1\right)^{k}x^{k-1}\zeta\left(k\right)=\gamma+\psi\left(x+1\right)$$ hence, assuming $x\neq0$ $$\sum_{k\geq2}\left(-1\right)^{k}x^{k-2}\zeta\left(k\right)=\frac{\gamma}{x}+\frac{\psi\left(x+1\right)}{x}$$ and now if we integrate from $0$ to $1$ $$\sum_{k\geq2}\frac{\left(-1\right)^{k}\zeta\left(k\right)}{k-1}=\int_{0}^{1}\frac{\gamma+\psi\left(x+1\right)}{x}dx$$ and there is no closed form, but only a numerical values ($1.257746...$). See for example here.

Hint: You might want first check if the integral $$\int_2^x \frac{1}{\xi} \log\left( 1+\frac{1}{\xi} \right) \mathrm d \xi$$ remains finite for $x\to\infty$.

Solution:

I got the limit $-\operatorname{Li}_2(-\frac12) = -\sum_{k=1}^\infty \frac{1}{(-2)^k k^2} \approx 0.45$. Not that satisfying...