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

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

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  • $\begingroup$ Please, write us your thoughts about the problem and what you have tried. $\endgroup$ – Blex Jul 4 '15 at 8:14
  • $\begingroup$ Notice, that logarithm is a continuous function. $\endgroup$ – luka5z Jul 4 '15 at 8:49
  • $\begingroup$ Yes i know that limit can go inside beacause of continuity of log... $\endgroup$ – user252347 Jul 4 '15 at 8:52
  • $\begingroup$ 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). $\endgroup$ – luka5z Jul 4 '15 at 8:53
  • $\begingroup$ also it is enough to calculate this product 2^(1/2).3^(1/6).4^(1/12).5^(1/20)............... $\endgroup$ – user252347 Jul 4 '15 at 8:53
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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.

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  • $\begingroup$ interesting.....Thanks $\endgroup$ – user252347 Jul 4 '15 at 9:05
  • $\begingroup$ As no closed form of the summation of $1/n^3$ is know, you can't really find a exact answer. $\endgroup$ – Ahmed S. Attaalla Jul 4 '15 at 9:11
  • $\begingroup$ @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 $\sum(\zeta(n)-1)/n=1-\gamma$. $\endgroup$ – Yves Daoust Jul 4 '15 at 9:14
  • $\begingroup$ 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 $1/n^3$. $\endgroup$ – Ahmed S. Attaalla Jul 4 '15 at 9:16
  • $\begingroup$ Or am I wrong @Yves Daoust $\endgroup$ – Ahmed S. Attaalla Jul 4 '15 at 9:19
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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.

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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...

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