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Solve the following limit: $$ \lim\limits_{n\to\infty}\prod\limits_{1\leq k\leq n}\left( 1+\frac{k}{n} \right)^{\frac{1}{k}} .$$

Here is what I do:

Take the logarithm: \begin{align} \lim\limits_{n\to\infty}\log \prod\limits_{1\leq k\leq n}\left( 1+\frac{k}{n} \right)^{\frac{1}{k}}&=\lim\limits_{n\to\infty}\sum_{k=1}^{n}\frac{\log (1+\frac{k}{n})}{\frac{k}{n}}\frac{1}{n}\\ &=\int_{0}^{1}\frac{\log (1+x)}{x}dx \end{align}

Then I am stuck. How to integrate $ \int_{0}^{1}\frac{\log (1+x)}{x}dx $?

The question is from: (8) of https://math.uchicago.edu/~min/GRE/files/week4.pdf

Edit: Different approaches are very welcome! The hint says 'estimate from above and below'. Maybe someone can provide a solution without integration?

Bach
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2 Answers2

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Find the Talyor series for $\log(1+x)/x$ and integrate termwise, and get $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}=\zeta(2)-\frac{2}{4}\zeta(2)=\frac{1}{2}\zeta(2)=\frac{\pi^2}{12}$$

Thomas Andrews
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As an alternative we have

$$\lim\limits_{n\to\infty}\log \prod\limits_{1\leq k\leq n}\left( 1+\frac{k}{n} \right)^{\frac{1}{k}}=\lim\limits_{n\to\infty}\sum_{k=1}^{n}\frac{\log (1+\frac{k}{n})}{k}$$

and for a fixed $k$

$$\frac{\log (1+\frac{k}{n})}{k}=\frac1k\sum_{j=1}^{\infty}\frac{(-1)^{j+1}(k/n)^j}{j}=\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{jn^j}k^{j-1}$$

and therefore by Faulhaber's formula

$$\sum_{k=1}^{n}\frac{\log (1+\frac{k}{n})}{k} =\sum_{j=1}^\infty \left[\frac{(-1)^{j+1}}{jn^j}\sum_{k=1}^{n}k^{j-1}\right] \to\sum_{j=1}^\infty \frac{(-1)^{j+1}}{{j^2}}=\frac{\pi^2}{12}$$

Refer also to the related

user
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