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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.

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    $\begingroup$ Can you find $\gamma$ such that $\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}} \sim \dfrac{c}{n^{\gamma}}$? $\endgroup$ – Daniel Fischer Mar 2 '16 at 21:18
  • $\begingroup$ I'm thinking maybe $\gamma=\frac{3}{2}$? $\endgroup$ – Superman Mar 2 '16 at 21:23
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    $\begingroup$ Right. (You should prove that.) Can you now see for which values of $\alpha$ the series converges? $\endgroup$ – Daniel Fischer Mar 2 '16 at 21:25
  • $\begingroup$ Yeah, $\alpha>\frac{2}{3}$. How should I prove that $\gamma=\frac{3}{2}$? $\endgroup$ – Superman Mar 2 '16 at 21:29
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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.

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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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