# Evaluating the $\lim_{n \to \infty} \prod_{1\leq k \leq n} (1+\frac{k}{n})^{1/k}$

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

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.

• The resulting integral is quite interesting and can be solved using the Taylor series for $\ln{(1+x)}$ to result in $$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$$ – Peter Foreman Jul 3 at 14:44

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. 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).

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

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 $$\lim_{x\to 0^+}\frac{\ln(1+x)}{x}=1$$.

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