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I am really struggling to work out the limit of the following product:

$$ \lim_{n \to \infty} \prod_{1\leq k \leq n} \left (1+\frac{k}{n} \right)^{1/k}.$$

So far, I have spent most of my time looking at the log of the above expression. If we set the desired limit equal to $L$, I end up with:

$$\log L = \lim_{n\to \infty}\log\left(\frac{n+1}{n} \right)+\frac{1}{2}\log\left(\frac{n+2}{n} \right) +\cdots +\frac{1}{n}\log\left(\frac{n+n}{n} \right),$$

which I can simplify to: $$ \log L = \lim_{n\to \infty} \log(n+1)+\frac{1}{2}\log(n+2)+\cdots \frac{1}{n}\log(2n)-\log(n)\left(1+\frac{1}{2}+\cdots\frac{1}{n}\right). $$

I tried to consider the above expression in a different form with an integral, but was unable to arrive at anything useful.

I have been stuck on this for quite awhile now, and would appreciate any insight.

Thanks

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Hint, based on Surb:
$$ \log L = \lim_{n\to\infty} \sum_{k=1}^n\frac{1}{k}\log\left(1+\frac{k}{n}\right) =\frac{1}{n}\sum_{k=1}^n\frac{n}{k}\log\left(1+\frac{k}{n}\right) =\int_0^1\frac{\log(1+x)}{x}\;dx $$
by a Riemann sum argument.

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    $\begingroup$ The resulting integral is quite interesting and can be solved using the Taylor series for $\ln{(1+x)}$ to result in $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$$ $\endgroup$ – Peter Foreman Jul 3 at 14:44
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I thought that it would be instructive to present an approach that does not rely on Riemann Sums, but rather makes use of the Taylor Series of $\log(1+x)$. To that end, we proceed.


The function $\log(1+x)$ can be represented by its Taylor Series, $\log(1+x)=\sum_{\ell=1}\frac{(-1)^{\ell-1}}{\ell}x^\ell$ for $-1<x\le 1$. Using this representation, we can write

$$\begin{align} \sum_{k=1}^n\log\left(1+\frac kn\right)^{1/k}&=\sum_{k=1}^n\frac1k\log\left(1+\frac kn\right)\\\\ &=\sum_{k=1}^n\left(\frac1k \sum_{\ell=1}^\infty \frac{(-1)^{\ell-1}}{\ell}\left(\frac kn\right)^\ell\right)\\\\ &=\sum_{\ell=1}^\infty \frac{(-1)^{\ell-1}}{\ell n^\ell}\sum_{k=1}^nk^{\ell-1}\tag1 \end{align}$$


Next, noting that $\displaystyle \sum_{k=1}^n k^{\ell-1}=\frac{n^\ell}{\ell}+O\left(n^{\ell -1}\right)$, we have from $(1)$ that

$$\begin{align} \sum_{k=1}^n\log\left(1+\frac kn\right)^{1/k}&=\sum_{\ell=1}^\infty \frac{(-1)^{\ell-1}}{\ell^2 }+O\left(\frac1n\right)\tag2 \end{align}$$


Finally, letting $n\to \infty$ in $(2)$ yields the result

$$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty \log\left(1+\frac kn\right)^{1/k}=\sum_{k=1}^\infty \frac{(-1)^{k=1}}{k^2}}\tag3$$


The series on the left-hand side of $(3)$ is equal to $\frac{\pi^2}{12}$ (See This).

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  • $\begingroup$ @will Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Jul 10 at 3:21
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hint

Taking logarithm, the product becomes

$$\frac 1n\sum_{k=1}^n\frac nk\ln(1+\frac kn)=$$

$$\frac 1n\sum_{k=1}^n\frac{\ln(1+\frac kn)}{\frac kn}$$

As a Riemann sum, the limit will be

$$\int_0^1\frac{\ln(1+x)}{x}dx$$

This improper integral is convergent since $\lim_{x\to 0^+}\frac{\ln(1+x)}{x}=1$.

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    $\begingroup$ The integral is NOT improper! The integrand has a removeable discontinuity. $\endgroup$ – Mark Viola Jul 3 at 15:02

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