Calculate the limit $\lim_{n\to\infty} \left(\prod_{k=1}^{n}(1+\frac{k}{n})\right)^{\frac{1}{n}}$

Question

I have to calculate this limit $$ \lim_{n\to\infty} \left(\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)\right)^{\frac{1}{n}}$$

I tried it just first to calculate the limit inside the product, but I think I got the answer 1. Any help?

Answer 1

If the limit be called $L$, then $$\log L= \lim_{n\to\infty} \frac 1n \sum_{k=1}^n \log\left(1+\frac kn\right) $$ Let $x=\frac kn$. Then this is equivalent to the integral $$\int_0^1 \log(1+x) \ dx $$ Applying by-parts, $$= x\log(1+x) \bigg|_0^1 -\int_0^1 \frac{x}{1+x} dx \\=\log 2-\int_0^1 dx+\int_0^1 \frac{dx}{1+x} \\=\log 2-1+\log 2\\=\log \frac 4e$$ And so $$L=\frac 4e$$

Answer 2

The Riemann integral approach is simple and elegant, but you can directly find it by Stirling's approximation:

$$n!\sim n^ne^{-n}\sqrt{2\pi n},$$

by simply writing your term as:

$$\left(\prod_{k=1}^n\dfrac{n+k}{n}\right)^{\frac 1n} = n^{-1}\sqrt[n]{\dfrac{(2n)!}{n!}}=\dots$$

Answer 3

As an alternative, we have that

$$\left(\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)\right)^{\frac{1}{n}}=e^{\frac{\sum_{k=1}^{n} \log\left(1+\frac{k}{n}\right) }{n}}=e^{\sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac12\frac{k^2}{n^3}+\frac13\frac{k^3}{n^4}+\ldots\right) }\to \frac4e$$

indeed by Faulhaber's formula

$$\sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac12\frac{k^2}{n^3}+\frac13\frac{k^3}{n^4}+\ldots\right)=\sum_{k=1}^n \frac{(-1)^{(k+1)}}{k(k+1)}+O\left(\frac1n\right)\to \ln 4-1$$

indeed by alternating harmonic series

$$\sum_{k=1}^n \frac{(-1)^{(k+1)}}{k(k+1)}=\sum_{k=1}^n \frac{(-1)^{(k+1)}}{k}-\sum_{k=1}^n \frac{(-1)^{(k+1)}}{k+1}\to\ln 2-(1-\ln 2)=2\ln 2-1$$

Answer 4

Let $$\displaystyle a_n = \prod_{k=1}^{n}\left(1+\frac{k}{n}\right)^{\frac{1}{n}}$$

Then, assuming that the limit exists, we know from this question that

$$\lim_{n\to\infty}{c_n}^{1/n} = \lim_{n\to\infty} \frac{c_{n+1}}{c_n}$$

Therefore

\begin{align} \lim_{n\to\infty}a_n&= \lim_{n\to\infty}\frac{\prod\limits_{k=1}^{n+1}\left(1+\frac{k}{n+1}\right)}{\prod\limits_{k=1}^{n}\left(1+\frac{k}{n}\right)}\\&= \lim_{n\to\infty}\frac{\prod\limits_{k=1}^{n+1}\left(n+1+k\right)}{\prod\limits_{k=1}^{n}\left(n+k\right)}\frac{\left(\frac{1}{n+1}\right)^{n+1}}{\left(\frac{1}{n}\right)^{n}}\\&= \lim_{n\to\infty}\frac{(n+2)\cdots(2n+2)}{(n+1)\cdots(2n)}\frac{\left(\frac{n}{n+1}\right)^n}{n+1}\\&=\lim_{n\to\infty}\frac{(2n+1)(2n+2)}{n+1}\frac{1}{\left(1+\frac1n\right)^n(n+1)}\\&=\lim_{n\to\infty}\frac{2(2n+1)}{n+1}\frac{1}{\left(1+\frac1n\right)^n}\\&= \frac{4}{e} \end{align}