Prove: $\lim_{n\to\infty} \left(\frac{n}{n+k}\right)^n=\frac{1}{e^k}$ 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.
 A: 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
$\to \frac1{e^k}$.
A: 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 $.
A: $$
\frac{n}{n+k}=\frac{(n+k)-k}{n+k}=1-\frac{k}{n+k}
$$
$$
\lim_{n\to\infty}(\frac{n}{n+k})^n=\lim_{n\to\infty}(1-\frac{k}{n+k})^n=\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k-k}
$$
$$
\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k-k}=\left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k}\right]\left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{-k}
\right]
$$
$$
\left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{-k}
\right]=1^{-k}=1
$$
$$
\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k-k}=\left[\lim_{n\to\infty}(1-\frac{k}{n+k})^{n+k}\right]\times1
$$
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
$$
\lim_{n\to\infty}(\frac{n}{n+k})^n=e^{-k}
$$
A: Hint, assuming it is known that $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e\,$:
$$
\lim_{n\to\infty} \left(\frac{n}{n+k}\right)^n = \lim_{n\to\infty} \left(\frac{1}{1+\frac{k}{n}}\right)^n= \frac{1}{\left(\lim_{n\to\infty} \left(1+\cfrac{\;1\;}{\frac{n}{k}}\right)^{\frac{n}{k}}\right)^k}= \frac{1}{e^k}
$$
A: $$a_n=\left(\frac{n}{n+k}\right)^n\implies \log(a_n)=n\log\left(\frac{n}{n+k}\right)=n \log\left(\frac{1}{1+\frac kn}\right)=-n\log\left({1+\frac kn}\right)$$ 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.
A: 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:
$$\lim_{n\to\infty}\left[\log{\left(\frac{n}{n+k}\right)^n}\right]=\lim_{n\to\infty}\left[n\log{\left(\frac{n}{n+k}\right)}\right]=\lim_{n\to\infty}\left[\frac{\log{\left(\frac{n}{n+k}\right)}}{\frac{1}{n}}\right]
\\=\lim_{n\to\infty}\left[\frac{\frac{k}{(n+k)^2}}{-\frac{1}{n^2}}\right]=\lim_{n\to\infty}\left[\frac{-kn^2}{n^2+2kn+k^2}\right]=\lim_{n\to\infty}\left[\frac{-k}{1+\frac{2k}{n}+\frac{k^2}{n^2}}\right]=-k.$$
