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.)

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

 A: Hint: $\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1} = \frac{-2}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n-1}+\sqrt{n+1})(\sqrt{n}+\sqrt{n-1})}$
So, this term is for big n approximately $-2n^{-\frac{3}{2}}$
A: 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 $\alpha < \frac 1 2$
A: First hint: $$\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1} = \sqrt{n+1}-\sqrt{n}+\sqrt{n-1}-\sqrt{n}$$
First term:
$$\sqrt{n+1}-\sqrt{n} = (\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$
Second term:
$$\sqrt{n-1}-\sqrt{n} = (\sqrt{n-1}-\sqrt{n})\frac{\sqrt{n-1}+\sqrt{n}}{\sqrt{n-1}+\sqrt{n}} = \frac{-1}{\sqrt{n-1}+\sqrt{n}}$$
Plug in the first and the second result into the original term:
$$\begin{align}
\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{-1}{\sqrt{n-1}+\sqrt{n}}&=\frac{\sqrt{n-1}-\sqrt{n+1}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n-1}+\sqrt{n})}\\
&=\frac{-2}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n-1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n-1})} \end{align}
$$
Thus $\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1} = O\left(\frac{1}{\sqrt{n}^3}\right)$. This should help you...
A: Note that $\sqrt{n\pm 1}=n^{1/2}\left(1\pm \frac{1}{n}\right)^{1/2}=n^{1/2}\left(1\pm \frac{1}{2n}-\frac{1}{8n^2}+O(n^{-3})\right)$.
Thus, we have
$$\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}=n^{1/2}\left(-\frac{1}{4n^2}+O(n^{-3})\right)$$
so that 
$$n^{\alpha}\left(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\right)=-\frac14 n^{\alpha -3/2}+O(n^{\alpha-5/2})$$
Thus the series converges for $\alpha<1/2$ and diverges elsewhere.
