Prove: $\lim_{n\to\infty} \left(\frac{n}{n+k}\right)^n=\frac{1}{e^k}$

Question

The limit $$\lim_{n\to\infty} \left(\frac{n}{n+k}\right)^n=\frac{1}{e^k}$$ is key to a couple of questions I am doing. I am just confused as to how this is derived.

Answer 1

This just occurred to me. Don't know how new it is.

$(\frac{n}{n+k})^n =\prod_{i=1}^k (\frac{n+i-1}{n+i})^n $.

For each $i$,

$\begin{array}\\ (\frac{n+i-1}{n+i})^n &=(\frac{n+i-1}{n+i})^{n+i}(\frac{n+i}{n+i-1})^{i}\\ &=(\frac{n+i-1}{n+i})^{n+i}(1+\frac{1}{n+i-1})^{i}\\ &\to \frac1{e} \end{array} $

so the product of $k$ of these $\to \frac1{e^k}$.

Answer 2

hint

$$(\frac {n}{n+k})^n=(\frac {n+k}{n})^{-n}$$

$$=(1+k/n)^{-n}=e^{-n\ln (1+k/n)} $$

$$\displaystyle {=e^{-k.\boxed {\frac {\ln (1+k/n)}{k/n}}} }$$

now use $$\lim_{X\to 0}\boxed {\frac {\ln (1+X)}{X}}=1$$

with $X=k/n $ and $n\to +\infty $.

Answer 3

$$ \frac{n}{n+k}=\frac{(n+k)-k}{n+k}=1-\frac{k}{n+k} $$

$$ \lim_{n\to\infty}(\frac{n}{n+k})^n=\lim_{n\to\infty}(1-\frac{k}{n+k})^n=\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k-k} $$

$$ \lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k-k}=\left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k}\right]\left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{-k} \right] $$

$$ \left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{-k} \right]=1^{-k}=1 $$

$$ \lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k-k}=\left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k}\right]\times1 $$

Now I define $m=n+k$. Note that when $k$ is finite: $m\to\infty$ when $n\to\infty$.

$$ \left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k}\right]=\lim_{m\to\infty}(1+\frac{-k}{m})^m=e^{-k} $$

Putting this altogether I conclude

$$ \lim_{n\to\infty}(\frac{n}{n+k})^n=e^{-k} $$

Answer 4

Hint, assuming it is known that $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e\,$:

$$ \lim_{n\to\infty} \left(\frac{n}{n+k}\right)^n = \lim_{n\to\infty} \left(\frac{1}{1+\frac{k}{n}}\right)^n= \frac{1}{\left(\lim_{n\to\infty} \left(1+\cfrac{\;1\;}{\frac{n}{k}}\right)^{\frac{n}{k}}\right)^k}= \frac{1}{e^k} $$

Answer 5

$$a_n=\left(\frac{n}{n+k}\right)^n\implies \log(a_n)=n\log\left(\frac{n}{n+k}\right)=n \log\left(\frac{1}{1+\frac kn}\right)=-n\log\left({1+\frac kn}\right)$$ Now, using Taylor expansion $$\log\left({1+\frac kn}\right)=\frac{k}{n}-\frac{k^2}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(a_n)=-k+\frac{k^2}{2 n}+O\left(\frac{1}{n^2}\right)$$ Taylor again $$a_n=e^{\log(a_n)}=e^{-k}\left(1+\frac{k^2}{2 n}\right)+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

Answer 6

There is a different approach, which requires more knowledge though (namely, we need limit of functions, continuity and derivatives):

As $$\lim_{n\to\infty} \left(\frac{n}{n+k}\right)^n=\lim_{n\to\infty} e^{\log{\left(\frac{n}{n+k}\right)^n}}=e^{\lim_{n\to\infty}\log{\left(\frac{n}{n+k}\right)^n}}$$ it is enough to show that $$\lim_{n\to\infty}\left[\log{\left(\frac{n}{n+k}\right)^n}\right]=-k.$$ This follows from L'Hôpital's rule: $$\lim_{n\to\infty}\left[\log{\left(\frac{n}{n+k}\right)^n}\right]=\lim_{n\to\infty}\left[n\log{\left(\frac{n}{n+k}\right)}\right]=\lim_{n\to\infty}\left[\frac{\log{\left(\frac{n}{n+k}\right)}}{\frac{1}{n}}\right] \\=\lim_{n\to\infty}\left[\frac{\frac{k}{(n+k)^2}}{-\frac{1}{n^2}}\right]=\lim_{n\to\infty}\left[\frac{-kn^2}{n^2+2kn+k^2}\right]=\lim_{n\to\infty}\left[\frac{-k}{1+\frac{2k}{n}+\frac{k^2}{n^2}}\right]=-k.$$