TITLE: For what values $\alpha$ does the sum converge?
QUESTION [1 upvotes]: For what values $\alpha$ does the sum converge? 
$$\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}- \frac{1}{\sqrt{n+1}}\right)^{\alpha} $$
I have no idea how to start solving this problem. I'm thinking of comparing it with another sum but not sure how to find that sum. Any help is appreciated.

REPLY [1 votes]: Another approach is to use Taylor's Theorem for the function $f(1/n)=\left(1+\frac1n\right)^{-1/2}$ around $0$.  
First note the equality
$$n^{-1/2}-(n+1)^{-1/2}=n^{-1/2}\left(1-\left(1+\frac1n\right)^{-1/2}\right)$$
Then using Taylor's Theorem, there exist numbers $0<\xi_n<\frac1n$ and $0<\zeta_n<\frac1n$, such that 
$$\begin{align}
\left(1+\frac1n\right)^{-1/2}&=1- (1+\zeta_n)^{-3/2}\frac1{2n}\\\\
&=  1-\frac1{2n}+(1+\xi_n)^{-5/2}\frac3{8n^2} 
\end{align}$$
Thus, we obtain the bounds for $n\ge 1$
$$1-\frac{1}{2n}\le \left(1+\frac1n\right)^{-1/2} \le 1-\frac{1}{2^{5/2}\,n} \tag 1$$
Using $(1)$, we find that
$$\begin{align}
\frac{1}{2^{5/2}\,n} \le n^{-1/2}-(n+1)^{-1/2}\le \frac{1}{2n^{3/2}}
\end{align}$$
Therefore, we have
$$\frac1{2^{5\alpha/2}}\sum_{n=1}^\infty \frac{1}{n^{3\alpha/2}}\le \sum_{n=1}^\infty\left(n^{-1/2}-(n+1)^{-1/2}\right)^\alpha\le \frac1{2^\alpha}\sum_{n=1}^\infty \frac{1}{n^{3\alpha/2}}$$
which from the comparison test converges for $\alpha <\frac23$ and diverges otherwise.
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