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!
 A: 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.
A: We have that
$$\begin{align*}
n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{\left(n+1\right)}{n}\right)
&=n^2\left(\exp\left(n\ln\left(1+\frac{1}{n\left(n+2\right)}\right)\right)-1-\frac{1}{n}\right)\\
&=n^2\left(\exp\left(n\left(\frac{1}{n\left(n+2\right)}+o\left(\frac{1}{n^3}\right)\right)\right)-1-\frac{1}{n}\right)\\
&=n^2\left(\exp\left(\frac{1}{n+2}+o\left(\frac{1}{n^2}\right)\right)-1-\frac{1}{n}\right)\\
&=n^2\left(1+\frac{1}{n+2}+\frac{1}{2(n+2)^2}+o\left(\frac{1}{n^2}\right)-1-\frac{1}{n}\right)\\
&=n^2\left(\frac{1/n}{1+2/n}+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)-\frac{1}{n}\right)\\
&=n^2\left(\frac{1}{n}\left(1-\frac{2}{n}+o\left(\frac{1}{n}\right)\right)+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)-\frac{1}{n}\right)\\
&=n^2\left(-\frac{2}{n^2}+\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\\
\end{align*}=-\frac{3}{2}+o\left(1\right).$$
Hence your limit as $n\to+\infty$ is $-\frac{3}{2}$.
A: We can proceed as follows
\begin{align}
L &= \lim_{n \to \infty}n^{2}\left\{\left(1 + \frac{1}{n(n + 2)}\right)^{n} - \frac{n + 1}{n}\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left\{\left(\frac{(n + 1)^{2}}{n(n + 2)}\right)^{n} - \frac{n + 1}{n}\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left\{\exp\left(n\log\left(\frac{(n + 1)^{2}}{n(n + 2)}\right)\right) - \exp\log\frac{n + 1}{n}\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\cdot\frac{n + 1}{n}\left\{\exp\left(n\log\left(\frac{(n + 1)^{2}}{n(n + 2)}\right) - \log\frac{n + 1}{n}\right) - 1\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\dfrac{\exp\left(n\log\left(\dfrac{(n + 1)^{2}}{n(n + 2)}\right) - \log\dfrac{n + 1}{n}\right) - 1}{n\log\left(\dfrac{(n + 1)^{2}}{n(n + 2)}\right) - \log\dfrac{n + 1}{n}}\cdot\left(n\log\left(\dfrac{(n + 1)^{2}}{n(n + 2)}\right) - \log\dfrac{n + 1}{n}\right)\notag\\
&= \lim_{n \to \infty}n^{2}\left(n\log\left(\dfrac{(n + 1)^{2}}{n(n + 2)}\right) - \log\dfrac{n + 1}{n}\right)\notag\\
&= \lim_{n \to \infty}n^{2}\left\{n\log\left(1 + \frac{1}{n(n + 2)}\right) - \log\left(1 + \frac{1}{n}\right)\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left\{n\left(\frac{1}{n(n + 2)} - \frac{1}{2n^{2}(n + 2)^{2}} + o(1/n^{4})\right) - \left(\frac{1}{n} - \frac{1}{2n^{2}} + o(1/n^{2})\right)\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left\{\left(\frac{1}{n + 2} - \frac{1}{2n(n + 2)^{2}} + o(1/n^{3})\right) - \left(\frac{1}{n} - \frac{1}{2n^{2}} + o(1/n^{2})\right)\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left(\frac{1}{n + 2} - \frac{1}{n}\right) + \frac{1}{2} + o(1)\notag\\
&= -2 + \frac{1}{2}\notag\\
&= -\frac{3}{2}\notag
\end{align}
We have used the standard limit $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ which shows that
\begin{align}
F(n) &= n\log\left(1 + \frac{1}{n(n + 2)}\right) - \log\frac{n + 1}{n}\notag\\
&= \frac{1}{n + 2}\cdot n(n + 2)\log\left(1 + \frac{1}{n(n + 2)}\right) - \log\frac{n + 1}{n}\notag\\
&\to 0\cdot 1 - 0 = 0\notag
\end{align}
Another standard limit $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$ is used to get rid of the exponential function and finally we use Taylor expansion $$\log(1 + x) = x - \frac{x^{2}}{2} + o(x^{2})$$ when $x \to 0$.
