<s>[INST] 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?
 [/INST] $$
\begin{align}
&\sum_{n=1}^\infty \frac{\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{n^n\left(\sqrt{1+\frac1n} - \sqrt{1+\frac1{n^2}}\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{n^n\left(\left(1+\frac1{2n}-\frac1{8n^2}+O\!\left(\frac1{n^3}\right)\right) - \left(1+\frac1{2n^2}+O\!\left(\frac1{n^4}\right)\right)\right)^n}{n^2}x^n\\
&=\sum_{n=1}^\infty \frac{\left(\frac12-\frac5{8n}+O\!\left(\frac1{n^2}\right)\right)^n}{n^2}x^n\\[3pt]
&=\sum_{n=1}^\infty \frac{e^{-5/4}+O\!\left(\frac1n\right)}{2^nn^2}x^n
\end{align}
$$
Thus, the radius of convergence is $2$, and the series converges absolutely at the boundaries
</s>