How to solve the limit $\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{n+1}{n}\right)\right]$? Hi I got an examination at the school which was so arduous that I'm stumped.
This problem is the toughest for me :
$$\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{n+1}{n}\right)\right]$$
I could not any approaching in order to solve problem . I need help!

From the binomial theorem,
$$ \left(1+\frac{1}{n(n+2)}\right)^n = \sum\limits_{k=0}^{n}{{n\choose k}\frac{1}{n^k(n+2)^k}} = 1 + n\frac{1}{n(n+2)} + \frac{n(n-1)}{2}\frac{1}{n^2(n+2)^2} + \sum\limits_{k=3}^{n}{{n\choose k}\frac{1}{n^k(n+2)^k}}. $$
Notice that ${n\choose k}\le n^k$, and hence ${n\choose k}\frac{1}{n^k(n+2)^k}\le\frac{1}{(n+2)^k}$ for all $k$. Thus,
\begin{align} \sum\limits_{k=3}^{n}{{n\choose k}\frac{1}{n^k(n+2)^k}}\le\sum\limits_{k=3}^{n}{\frac{1}{(n+2)^k}} &= \frac{1}{(n+2)^3} + \sum\limits_{k=4}^{n}{\frac{1}{(n+2)^k}}\\
&\le \frac{1}{(n+2)^3}+(n-4)\frac{1}{(n+2)^4}\\
&\le \frac{2}{(n+2)^3}
\end{align}
so
$$\lim\limits_{n\rightarrow\infty}{n^2\sum\limits_{k=3}^{n}{{n\choose k}\frac{1}{n^k(n+2)^k}}}\le\lim\limits_{n\rightarrow\infty}{n^2\frac{2}{(n+2)^3}} = 0. $$
Hence, it is sufficient to calculate
$$\lim\limits_{n\rightarrow\infty}{n^2\left(\left(1+n\frac{1}{n(n+2)} + \frac{n(n-1)}{2}\frac{1}{n^2(n+2)^2}\right)-\frac{n+1}{n}\right)} $$
which should not be hard to do.
