Behaviour of $\sum\limits_{n=1}^\infty \frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n$ for $|x|=2$ I want to determine the radius of convergence for the power series
$$\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} x^n$$
and determine what happens at the boundaries. I determined the ratio of convergence $R$ to be $2$, but I struggle to show that it converges on $x = \pm 2$. Can you give me a hint?

What I have so far: $R = \frac{1}{\limsup_{n \to \infty} (x_n)} = \frac{1}{\frac{1}{1+1}} = \frac{2}{1} = 2$, where
$$\begin{align} x_n := \sqrt[n]{|a_n|} &= \sqrt[n]{| \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2}|} = \frac{\sqrt{n^2+n} - \sqrt{n^2+1}}{1} \\ &= \frac{(n^2+n) - (n^2+1)}{\sqrt{n^2+n} + \sqrt{n^2+1}} = \frac{(n^2+n) - (n^2+1)}{\sqrt{n^2+n} + \sqrt{n^2+1}} \\ &= \frac{n - 1}{\sqrt{n^2+n} + \sqrt{n^2+1}} = \frac{1 - \frac{1}{n}}{\sqrt{1+\frac{1}{n}} + \sqrt{1+\frac{1}{n^2}}} \end{align}$$
How do I determine if $\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} (\pm 2)^n$ converges? The root test doesn't work and the ratio test doesn't seem to be a smart move. What can I do?
 A: Let's try to bound the sequence here after we call it $a_n$. Note that the following inequalities hold:
$$a_n={{(\sqrt{n^2+n}-\sqrt{n^2+1})^n}\over{n^2}}x^n$$$$={{(n-1)^n}\over{n^2}(\sqrt{n^2+n}+\sqrt{n^2+1})^n}x^n$$$$\le {{(n-1)^n}\over{n^2}(2n)^n}x^n$$$$\le{{n^n}\over{n^2}(2n)^n}={{({x\over 2})^n}\over{n^2}}=b_n$$
where $b_n$ obviously converges when $|{x\over 2}|\le 1$ or $|x|\le 2$. 
Now we try to bound $a_n$ from bottom. We have :
$$a_n\ge {{(n-1)^n}\over{n^2}(\sqrt{n^2+n+{1\over 4}}+\sqrt{n^2+n+{1\over 4}})^n}x^n={{(n-1)^n}\over{n^2}(2n+1)^n}x^n={1\over {n^2}}({{n-1}\over{2n+1}})^nx^n={1\over {n^22^n}}(1-{3\over{2n+1}})^{{{2n}\over 3}{3\over 2}}x^n$$
Now note that the term $(1-{3\over{2n+1}})^{{{2n}\over 3}{3\over 2}}$ tends to $e^{3\over2}$ which is a constant so 
$$a_n=\Theta({1\over {n^2}}({x\over 2})^n)$$
 which diverges when $|x|>2$ so is $a_n$. Therefore we finally deduce that the radius of convergence is $2$.
A: Note that
$$\sqrt{n^2+n}=n\left(1+\frac1n\right)^{\frac12}\sim n+\frac12 \quad \sqrt{n^2+1}=n\left(1+\frac1{n^2}\right)^{\frac12}\sim n$$
$$\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n\sim\frac{1}{2^n}$$
thus
$$\frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n \sim \frac{x^n}{2^nn^2}$$
which diverges when $|x|>2$.
A: $$
\begin{align}
&\sum_{n=1}^\infty \frac{\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{n^n\left(\sqrt{1+\frac1n} - \sqrt{1+\frac1{n^2}}\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{n^n\left(\left(1+\frac1{2n}-\frac1{8n^2}+O\!\left(\frac1{n^3}\right)\right) - \left(1+\frac1{2n^2}+O\!\left(\frac1{n^4}\right)\right)\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{\left(\frac12-\frac5{8n}+O\!\left(\frac1{n^2}\right)\right)^n}{n^2}x^n\\[3pt]
&=\sum_{n=1}^\infty \frac{e^{-5/4}+O\!\left(\frac1n\right)}{2^nn^2}x^n
\end{align}
$$
Thus, the radius of convergence is $2$, and the series converges absolutely at the boundaries
