Limit of a product $(1 + \frac{k}{n})$

Question

I am trying to find the limit of

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

I have that it is equivalent to the

$$\lim_{n\to\infty}\frac{1}{n}\left(\sum_{k=1}^n\ln\left(1+\frac{k}{n}\right)\right)$$

I can't get any further than this. Any help would be much appreciated.

Careful, the limits aren't the same. – vonbrand Sep 20 '15 at 14:48 — vonbrand, Sep 20 '15 at 14:48
The second is the natural logarithm of the first. – Akiva Weinberger Sep 20 '15 at 16:05 — Akiva Weinberger, Sep 20 '15 at 16:05

score 12 · Accepted Answer · answered Sep 20 '15 at 14:42

12

You have a Riemann sum: $$ \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)=\int_{0}^{1}\log(1+x)\,dx = \color{red}{-1+2\log 2},$$ hence the original limit equals $\large\color{red}{\frac{4}{e}}$.

answered Sep 20 '15 at 14:42

Jack D'Aurizio

353,855

juantheron · Answer 2 · 2015-09-20T14:53:30.653

Let $$\displaystyle L = \lim_{n\to\infty}\left(\prod_{k=1}^n\left(1+\frac{k}{n}\right)\right)^{1/n} = \lim_{n\rightarrow \infty}\left[\left(\frac{n+1}{n}\right)\cdot \left(\frac{n+2}{n}\right)....\left(\frac{n+n}{n}\right)\right]^{\frac{1}{n}}$$

So we get $$\displaystyle L = \lim_{n\to\infty}\left[\frac{(2n)!}{n^n\cdot n!}\right]^{\frac{1}{n}}$$

Now Using Stirling approximation $$\displaystyle n! = \left(\frac{n}{e}\right)^{n}\cdot \sqrt{2\pi n}$$

So we get $$\displaystyle L = \lim_{n\to\infty}\left[\frac{\left(\frac{2n}{e}\right)^{2n}\cdot \sqrt{4\pi n}}{n^n\cdot \left(\frac{n}{e}\right)^{n}\cdot \sqrt{2\pi n}}\right]^{\frac{1}{n}}=\lim_{n\rightarrow \infty}\left[\frac{\frac{4n^2}{e^2}\times \sqrt{2}}{\frac{n^2}{e}}\right]^{\frac{1}{n}}$$

So we get $$\displaystyle L = \frac{4}{e}\cdot \lim_{n\rightarrow \infty}(2)^{\frac{1}{2n}}= \frac{4}{e}.$$

You used the Stirling approximation, how can one conclude that the limit is really the same? I think you need to use the O-Notation for the remainder. — MrYouMath, Sep 20 '15 at 15:11

score 2 · Answer 3 · answered Sep 20 '15 at 14:51

Note that $$ \sum_{k\leq n}\log\left(1+\frac{k}{n}\right)=\sum_{k\leq n}\log\left(k+n\right)-n\log\left(n\right) $$ and using Abel's summation formula we have $$\sum_{k\leq n}\log\left(k+n\right)=\sum_{n<k\leq2n}\log\left(k\right)=2n\log\left(2n\right)-n\log\left(n\right)-\int_{n}^{2n}\frac{\left\lfloor t\right\rfloor }{t}dt $$ where $\left\lfloor t\right\rfloor $ is the floor function. So, using $\left\lfloor t\right\rfloor =t+O\left(1\right) $ we get $$=n\log\left(n\right)+2n\log\left(2\right)-n+O\left(\log\left(n\right)\right) $$ hence $$\frac{1}{n}\sum_{k\leq n}\log\left(1+\frac{k}{n}\right)=\frac{1}{n}\left(2n\log\left(2\right)-n+O\left(\log\left(n\right)\right)\right)\rightarrow2\log\left(2\right)-1. $$

Limit of a product $(1 + \frac{k}{n})$

3 Answers3