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.
I think the following can be a good start (or perhaps an almost complete answer. I need to go now)
$$\frac1{\sqrt n}-\frac1{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt n}{\sqrt{n^2+n}}=$$
$$\frac1{\sqrt{n^3+n^2}+\sqrt{n^3+2n^2+n}}\le\frac1{2\sqrt{n^3}}=\frac1{2n^{3/2}}$$
Thus, if $\;\frac32\alpha>1\implies\alpha>\frac23\;$, the series converges by the comparison test
Observe that if $\;\alpha=\frac23\;$ , then
$$\left(\frac1{\sqrt n}-\frac1{\sqrt{n+1}}\right)^{2/3}=\left(\frac{\sqrt{n+1}-\sqrt n}{\sqrt{n^2+n}}\right)^{2/3}=$$
$$\left(\frac1{\sqrt{n^3+n^2}+\sqrt{n^3+2n^2+n}}\right)^{2/3}\ge\left(\frac1{2\sqrt{4n^3}}\right)^{2/3}=\frac1{2^{5/3}\sqrt2}\cdot\frac1{n}$$
and thus the series diverges.
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.
