Computing a limit similar to the exponential function I want to show the following limit:
$$
\lim_{n \to \infty}
n
\left[
    \left( 1 - \frac{1}{n} \right)^{2n}
    - \left( 1 - \frac{2}{n} \right)^{n}
\right]
    = \frac{1}{e^{2}}.
$$
I got the answer using WolframAlpha, and it seems to be correct numerically, but I am having trouble proving the result. My first instinct was to write the limit as
$$
\lim_{n \to \infty}
\frac
{
    \left( 1 - \frac{1}{n} \right)^{2n}
    - \left( 1 - \frac{2}{n} \right)^{n}
}
{1/n}.
$$
Then, I tried applying l'Hopital's rule, and I got
$$
\lim_{n \to \infty}
\frac
{
    \left( 1 - \frac{1}{n} \right)^{2n}
    \left( 2 \log\left( 1 - \frac{1}{n} \right) + \frac{2}{n-1} \right)
    -
    \left( 1 - \frac{2}{n} \right)^{n}
    \left( \log\left( 1 - \frac{2}{n} \right) + \frac{2}{n-2} \right)
}
{-1/n^{2}}.
$$
This does not seem to have gotten me anywhere. My second attempt was to use the binomial theorem:
$$
\begin{align*}
n
\left[
    \left( 1 - \frac{1}{n} \right)^{2n}
  - \left( 1 - \frac{2}{n} \right)^{n}
\right]
&
=
n
\left[
    \sum_{k=0}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k}}
  - \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^{k} 2^{k}}{n^{k}}
\right]
\\ &
=
\sum_{k=2}^{n} \left[ \binom{2n}{k} - \binom{n}{k} 2^{k} \right] \frac{(-1)^{k}}{n^{k-1}}
+ \sum_{k=n+1}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k-1}}.
\end{align*}
$$
At this point I got stuck again.
 A: I would do the transform 
$$\left(1-\frac{1}{n}\right)^{2n}=e^{2n\log\left(1-\frac{1}{n}\right)}$$
then use the second order Taylor expansion
$$\log(1+x)\approx x-\frac{x^2}{2}$$
and similarly for the other term, obtaining
$$
n\left(e^{-2-\frac{1}{n}}-e^{-2-\frac{2}{n}}\right)=e^{-2-\frac{2}{n}}\cdot\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}
$$
A: We can proceed in the following manner
\begin{align}
L &= \lim_{n \to \infty}n\left\{\left(1 - \frac{1}{n}\right)^{2n} - \left(1 - \frac{2}{n}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}n\left(1 - \frac{2}{n}\right)^{n}\left\{\dfrac{\left(1 - \dfrac{1}{n}\right)^{2n}}{\left(1 - \dfrac{2}{n}\right)^{n}} - 1\right\}\notag\\
&= \frac{1}{e^{2}}\lim_{n \to \infty}n\left\{\dfrac{\left(1 - \dfrac{2}{n} + \dfrac{1}{n^{2}}\right)^{n}}{\left(1 - \dfrac{2}{n}\right)^{n}} - 1\right\}\notag\\
&= \frac{1}{e^{2}}\lim_{n \to \infty}n\left\{\left(1 + \frac{1}{n^{2} - 2n}\right)^{n} - 1\right\}\notag\\
&= \frac{1}{e^{2}}\lim_{n \to \infty}f(n)\tag{1}
\end{align}
Now note that 
\begin{align}
n\left(1 + \frac{n}{n^{2} - 2n} - 1\right) < f(n) &= n\left[\left\{\left(1 + \frac{1}{n^{2} - 2n}\right)^{n^{2} - 2n}\right\}^{n/(n^{2} - 2n)} - 1\right]\notag\\
&< n\left\{\exp\left(\frac{n}{n^{2} - 2n}\right) - 1\right\}\notag\\
&= n\cdot\frac{n}{n^{2} - 2n}\cdot\frac{n^{2} - 2n}{n}\left\{\exp\left(\frac{n}{n^{2} - 2n}\right) - 1\right\}\notag\\
\end{align}
and we get $$\frac{n^{2}}{n^{2} - 2n} < f(n) < \frac{n^{2}}{n^{2} - 2n}\cdot\frac{n^{2} - 2n}{n}\left\{\exp\left(\frac{n}{n^{2} - 2n}\right) - 1\right\}$$ Taking limits as $n \to \infty$ and using squeeze theorem we see that $f(n) \to 1$ as $n \to \infty$. From equation $(1)$ we can see that $L = 1/e^{2}$.
In order to derive the inequalities we have used $(1 + x)^{n} > 1 + nx$ for $x > 0$ and $n$ a positive integer and $(1 + (1/n))^{n} < e$ for all positive integers $n$. Both these inequalities are pretty standard and can be easily proved. Also note that if we put $t = n/(n^{2} - 2n)$ then $t \to 0$ as $n \to \infty$ and hence $(e^{t} - 1)/t \to 1$ as $n \to \infty$.
A: $$
(1-\frac1{n})^{2n} =(1-\frac2{n}+\frac1{n^2})^n
$$
use the binomial expansion:
$$
(1-\frac2{n}+\frac1{n^2})^n =\sum_{k=0}^n(1-\frac2{n})^{n-k}\binom{n}{k}\frac1{n^{2k}}
$$
so
$$
n
\left[
    \left( 1 - \frac{1}{n} \right)^{2n}
    - \left( 1 - \frac{2}{n} \right)^{n}
\right] = \sum_{k=1}^n(1-\frac2{n})^{n-k}\binom{n}{k}\frac1{n^{2k-1}} \\
=(1-\frac2{n})^{n-1}+\sum_{k=2}^n(1-\frac2{n})^{n-k}\binom{n}{k}\frac1{n^{2k-1}} 
$$
A: In the same spirit as Lucian, you can consider the general expansion for large values of $n$ $$ \left( 1 - \frac{a}{n} \right)^{b\,n}= \left(1-\frac{a^2 b}{2 n}+\frac{a^3 b (3 a b-8)}{24 n^2}-\frac{a^4 b
   \left(a^2 b^2-8 a b+12\right)}{48 n^3}+\cdots\right)e^{-a b} $$ Applied to your case $$\left( 1 - \frac{1}{n} \right)^{2n}=\left(1-\frac 1n-\frac1 {6n^2}+\cdots\right)e^{-2}$$ $$\left( 1 - \frac{2}{n} \right)^{n}=\left(1-\frac 2n-\frac2 {3n^2}+\cdots\right)e^{-2}$$ which finally make $$n
\left[
    \left( 1 - \frac{1}{n} \right)^{2n}
    - \left( 1 - \frac{2}{n} \right)^{n}
\right]=\left( 1 +\frac{1}{2n}+\cdots \right)e^{-2}$$
In a more general manner,$$\left[
    \left( 1 - \frac{a}{n} \right)^{bn}
    - \left( 1 - \frac{b}{n} \right)^{an}
\right]=(a-b)\left(-\frac{a b }{2 n}+\frac{a b  (a+b) (3 a b-8)}{24 n^2}\right)e^{-ab}$$
