For which values of $\alpha \in \mathbb{R}$, does the series $\sum_{n=1}^\infty n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1})$ converge? How do I study for which values of $\alpha \in \mathbb{R}$ the following series converges?
(I have some troubles because of the form [$\infty - \infty$] that arises when taking the limit.)

$$\sum_{n=1}^\infty n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1})$$


Using Taylor expansion $\sqrt{1+x} =_{x\to 0} 1 + \frac x 2 - \frac {x^2}{8} + O(x^3)$, we obtain for $n \to \infty$:
$$\begin{eqnarray}
n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1}) & = & n^\alpha \sqrt{n} \left(\sqrt{1+\frac 1 n} - 2 + \sqrt{1-\frac 1 n}\right) \\
& = & n^{\alpha + 1/2} \left(1 + \frac 1 {2n} - \frac 1 {8n^2} - 2 + 1 - \frac {1}{2n} - \frac 1 {8n^2} + O(\frac 1 {n^3}) \right) \\
& = & -n^{\alpha + 1/2} \left(\frac 1 {4n^2}  + O(\frac 1 {n^3})\right) \\
& \sim_{n\to\infty} & \frac{-1}{4 n^{3/2 - \alpha}}
\end{eqnarray}$$
Therefore, the series converges if and only if $\frac 3 2 - \alpha > 1$, ie $\alpha < \frac 1 2$
