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?

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 $|x|>2$.
