Compute $\lim\limits_{n\to\infty}n^2\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^n\right)$ 
A problem I saw recently is to compute $$\lim_{n\to\infty} n^2 \left(\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\right)$$

I thought it would be fun to give this a go but it completely stumped me. I am not aware of any of the standard techniques used to compute $\infty\times (e-e)$ - type limits, bar possibly L'Hôpital's Rule. But I doubt that would be a good idea here, at best it would be ugly. I would appreciate solutions using any methods, the more elegant the better of course. Though I welcome any relevant hints as well if full solutions are discouraged.
 A: Let consider the sequence:
\begin{align}
x_n
&=n\log\left(1+\frac 1n\right)\\
&\sim n\left(\frac 1n-\frac 1{n^2}+o\left(\frac 1{n^2}\right)\right)\\
&\sim 1-\frac 1{2n}+\frac 1{3n^2}+o\left(\frac 1{n^2}\right)
\end{align}
Then
\begin{align}
x_{n+1}-x_n
&=\left(1-\frac 1{2(n+1)}+\frac 1{3(n+1)^2}\right)-\left(1-\frac 1{2n}+\frac 1{3n^2}\right)+o\left(\frac 1{n^2}\right)\\
&=\frac 1{2n(n+1)}+\frac 1{3(n+1)^2}-\frac 1{3n^2}+o\left(\frac 1{n^2}\right)\\
&=\frac 1{2n^2}+o\left(\frac 1{n^2}\right)\\
&\sim\frac 1{2n^2}
\end{align}
consequently
\begin{align}
n^2 \left(\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\right)
&=n^2(e^{x_{n+1}}-e^{x_n})\\
&=n^2e^{x_n}(e^{x_{n+1}-x_n}-1)\\
&\sim n^2e^{x_n}(x_{n+1}-x_n)\\
&\sim \frac{e^{x_n}}2\\
&\to\frac e2
\end{align}
A: Too long for a comment.
Using the same approach as in Fabio Lucchini's answer, you could even get more than the limit itself at the price of one single extra term in the expansions.
$$x_n=n\log\left(1+\frac 1n\right)=1-\frac{1}{2 n}+\frac{1}{3 n^2}-\frac{1}{4
   n^3}+O\left(\frac{1}{n^4}\right)$$
$$e^{x_n}=e-\frac{e}{2 n}+\frac{11 e}{24 n^2}-\frac{7 e}{16
   n^3}+O\left(\frac{1}{n^4}\right)$$ Doing the same and continuing with Taylor series (or long division)
$$x_{n+1}=(n+1)\log\left(1+\frac 1{n+1}\right)=1-\frac{1}{2 n}+\frac{5}{6 n^2}-\frac{17}{12
   n^3}+O\left(\frac{1}{n^4}\right)$$
$$e^{x_{n+1}}=e-\frac{e}{2 n}+\frac{23 e}{24 n^2}-\frac{89 e}{48
   n^3}+O\left(\frac{1}{n^4}\right)$$
$$e^{x_{n+1}}-e^{x_{n}}=\frac{e}{2 n^2}-\frac{17 e}{12 n^3}+O\left(\frac{1}{n^4}\right)$$
$$n^2\left(e^{x_{n+1}}-e^{x_{n}} \right)=\frac{e}{2}-\frac{17 e}{12 n}+O\left(\frac{1}{n^2}\right)$$ which shows th limpt and also how it is approached.
Try even with a small number such as $n=10$. The exact result would be $\frac{29833762621839898789}{28531167061100000000}$ which is $\approx 1.04566$ while the above truncated expression would give $\frac{43 e}{120}\approx 0.97405$.
A: Hint: Let the function be $f(n)$. This function has a horizontal asymptote as $n\to\infty$; the answer is $e/2$. So, we're left with actually proving the asymptote exists (i.e. $f'\to0^+$ and $f\not\to\infty$) and that the asymptote occurs at $y=e/2$.
A: The expression under limit is of the form $n^2(A-B)$ where $A, B$ tend to $e$ as $n\to\infty $. We can write $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A-\log B} \cdot (\log A-\log B) $$ and the fraction in middle tends to $1$ so that the desired limit is equal to the limit of the expression $en^2(\log A-\log B) $. Now we can see that $$\log A-\log B =(n+1)\log\left(\frac{n+2}{n+1}\right)-n\log\left(\frac{n+1}{n}\right)$$ which can be written as $$n\log\frac{n^2+2n}{n^2+2n+1}+\log\frac{n+2}{n+1}$$ or $$(n+1)\log\left(1-\frac{1}{(n+1)^2}\right)+\log\left(1+\frac{1}{n+1}\right)-\log\left(1-\frac{1}{(n+1)^2}\right)$$ The last term when multiplied by $n^2$ tends to $-1$ and hence the desired limit is equal to limit of $$en^2\left\{(n+1)\log\left(1-\frac{1}{(n+1)^2}\right)+\log\left(1+\frac{1}{n+1}\right)\right\}+e$$ which is same as that of $$en^2\left\{n\log\left(1-\frac{1}{n^2}\right)+\log\left(1+\frac{1}{n}\right)\right\}+e\tag{1}$$ Using Taylor series or L'Hospital's Rule we can easily see that $$\lim_{x\to 0}\frac{\log(1+x)-x}{x^2}=-\frac{1}{2}$$ Putting $x=1/n$ and $-1/n^2$ respectively we get $$\lim_{n\to\infty} n^2\left(\log\left(1+\frac{1}{n}\right)-\frac{1}{n}\right)=-\frac{1}{2}\tag{2}$$ and $$\lim_{n\to\infty} n^3\left(\log\left(1-\frac{1}{n^2}\right)+\frac{1}{n^2}\right)=0\tag{3}$$ Using $(2),(3)$ we see that the limit of expression in $(1)$ is $e/2$.
A: Using only the Binomial Theorem and Geometric Series Formula:
$$
\begin{align}
&n^2\left(\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^n\right)\tag1\\
&=n^2\left(1+\frac1n\right)^{n+1}\left(\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}-\frac{n}{n+1}\right)\tag2\\
&=n^2\left(1+\frac1n\right)^{n+1}\left(\left(1-\frac1{(n+1)^2}\right)^{n+1}-\left(1-\frac1{n+1}\right)\right)\tag3\\
&=n^2\left(1+\frac1n\right)^{n+1}\left({\scriptsize\left(\color{#C00}{1-\frac1{n+1}}+\color{#090}{\frac{n}{2(n+1)^3}}+\color{#00F}{\sum_{k=3}^{n+1}\binom{n+1}{k}\frac{(-1)^k}{(n+1)^{2k}}}\right)-\left(\color{#C00}{1-\frac1{n+1}}\right)}\right)\tag4\\
&=n^2\left(1+\frac1n\right)^{n+1}\left(\color{#090}{\frac{n}{2(n+1)^3}}+\color{#00F}{O\!\left(\frac1{n^3}\right)}\right)\tag5\\
&=\left(1+\frac1n\right)^{n+1}\left(\frac{n^3}{2(n+1)^3}+O\!\left(\frac1n\right)\right)\tag6\\[6pt]
&\to\frac e2\tag7
\end{align}
$$
Explanation:
$(1)$: quantity of which to find the limit as $n\to\infty$
$(2)$: pull out a factor from both terms in the difference
$(3)$: rewrite both minuend and subtrahend as a difference from $1$
$(4)$: apply the binomial theorem
$(5)$: cancel the red terms and note that
$\phantom{\text{(5):}}$ $\sum\limits_{k=3}^{n+1}\binom{n+1}{k}\frac1{(n+1)^{2k}}\le\sum\limits_{k=3}^{n+1}\frac1{k!}\frac1{(n+1)^k}\le\sum\limits_{k=3}^\infty\frac1{6\cdot4^{k-3}}\frac1{(n+1)^3}=\frac2{9(n+1)^3}$
$(6)$: distribute $n^2$ over the difference
$(7)$: limit of a product is the product of the limits
A: 
Only necessary ingredient: $\log(1+x)=x-\frac12x^2+o(x^2)$ when $x\to0$; in particular, $\frac1x\log(1+x)\to1$ when $x\to0$.

Rewrite this as $$x_n=n^2\left(f\left(\frac1{n+1}\right)-f\left(\frac1n\right)\right)$$ where $$f(x)=\left(1+x\right)^{1/x}$$ Now, $f(x)=\exp\left(\frac1x\log\left(1+x\right)\right)$ hence $$f'(x)=g(x)\frac1{x^2(1+x)}f(x)$$ with 
$$g(x)=x-(1+x)\log(1+x)$$
When $x\to0$, $\log(1+x)=x-\frac12x^2+o(x^2)$ hence $$g(x)=x-(1+x)\left(x-\frac12x^2+o(x^2)\right)\sim-\frac12x^2$$ Since $f(x)\to e$ when $x\to0$, this yields $$f'(x)\sim-\frac12x^2\cdot\frac1{x^2}\cdot e=-\frac12e$$ hence $$f\left(\frac1{n+1}\right)-f\left(\frac1n\right)\sim-\frac12e\left(\frac1{n+1}-\frac1n\right)\sim\frac12e\frac1{n^2}$$ and finally $$\lim x_n=\frac12e$$
