Does $a_n:=\left(1+\frac1{n+c}\right)^n$ converge to $e$, for $c\in\Bbb{N}$? In my lecture, we defined $e$ as
$$\lim_{n\to\infty}\left(1+\frac 1n\right)^n = \sum_{k=0}^\infty \frac 1{k!} = e$$
While playing around with those definitions, I stumbled onto
$$a_n := \left(1 + \frac{1}{n +c}\right)^n$$ where $c \in \mathbb{N}$. From plotting it, it seems like it converges to $e$ as well, but slower if $c$ gets larger, but I can't find a proof for it.
Is my assumption true, and if so how would I go about proving it?
 A: Consider:
$$\lim_{n\to\infty}\bigg(1 + \frac{1}{n+c}\bigg)^{n}=\lim_{n\to\infty}\frac{\displaystyle\bigg(1 + \frac{1}{n + c}\bigg)^{n+ c}}{\displaystyle\bigg(1 + \frac{1}{n+c}\bigg)^{c}}=\frac{\displaystyle\lim_{n\to\infty}\bigg(1 + \frac{1}{n + c}\bigg)^{n + c}}{
\displaystyle\lim_{n\to\infty}\bigg(1 + \frac{1}{n + c}\bigg)^{c}}$$
Because $n + c\to\infty$ as $n\to\infty$:
$$=\frac{e}{\displaystyle\lim_{n\to\infty}\bigg(1+\frac{1}{n + c}\bigg)^{c}} = \frac{e}{1^{c}}=\boxed{e}$$
A: You can show this with a quick change of variables.  Let $m = n + c$, which means that $$a_{n} = \left(1 + \frac{1}{m}\right)^{m-c} = \left(1 + \frac{1}{m}\right)^{m}\left(1 + \frac{1}{m}\right)^{-c}$$  Now, as $m\to\infty$ as $n\to\infty$, we have
\begin{align}
\lim_{n\to\infty}a_{n} &= \lim_{m\to\infty}\left(1 + \frac{1}{m}\right)^{m}\left(1 + \frac{1}{m}\right)^{-c}\\
&= \lim_{m\to\infty}\left(1 + \frac{1}{m}\right)^{m}\cdot\lim_{m\to\infty}\left(1 + \frac{1}{m}\right)^{-c}\\
&=e\cdot 1\\
&=e
\end{align}
A: $$a_n=\left(1+\frac1{n+c}\right)^n\implies \log(a_n)=n \log\left(1+\frac1{n+c}\right)$$
$$\log\left(1+\frac1{n+c}\right)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\frac 1{(n+c)^k}=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\frac 1{n^k}\frac 1{\left(1+\frac{c}{n}\right)^k }$$
$$\log\left(1+\frac1{n+c}\right)=\sum_{k=1}^\infty\frac{(-c)^k-(-c-1)^k}{k}\frac 1{n^k}$$ Computing the first terms
$$\log\left(1+\frac1{n+c}\right)=\frac{1}{n}-\frac{2c+1}{2n^2}+\frac{3c^2+3c+1}{3n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(a_n)=1-\frac{2c+1}{2n}+\frac{3c^2+3c+1}{3n^2}+O\left(\frac{1}{n^3}\right)$$
$$a_n=e^{\log(a_n)}=e \left(1-\frac{2 c+1}{2 n}+\frac{36 c^2+36c +11}{24
   n^2}\right)+O\left(\frac{1}{n^3}\right)$$
