# Convergence of parameterized series $\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} - \sqrt{n})^p \cdot \ln\left( \frac{n-1}{n+1}\right) \right)$

$$\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} - \sqrt{n})^p \cdot \ln\left( \frac{n-1}{n+1}\right) \right)$$

I guess that more useful form is:

$$\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} + \sqrt{n})^{-p} \cdot \left( \ln(n-1) - \ln(n+1) \right) \right)$$

The question is, for what $p \in \mathbb{R}$ does it converge?

I've made some experimental random checks, and there seems that there's something happening for -1 and -2, I think it converges for $p \in (-2; -1) \cup (-1; +\infty)$, but then again it's just some random test results.

How could I find for which $p$ it works, and which method of examining convergence should I use?

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HINT: Rewrite the series as $$\begin{eqnarray} \mathcal{S} &=& \sum_{n=2}^\infty \left( \sqrt{n+1}-\sqrt{n} \right)^p \cdot \ln \left(\frac{n-1}{n+1} \right) \\ &=& \sum_{n=2}^\infty \left( \frac{\sqrt{n+1}^2-\sqrt{n}^2}{\sqrt{n+1}+\sqrt{n}} \right)^p \cdot \ln \left(1-\frac{2}{n+1} \right)\\ &=& \sum_{n=2}^\infty \frac{ \ln \left(1-\frac{2}{n+1} \right)}{\left(\sqrt{n+1} + \sqrt{n} \right)^p} \end{eqnarray}$$

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Writing it as

$$\sum\limits_{n = 2}^{ + \infty } {\left( {{{\left( {\sqrt {n + 1} - \sqrt n } \right)}^p}\cdot\ln \left( {1 - {2 \over {n + 1}}} \right)} \right)}$$

and using $$\log \left( {1 - {2 \over {n + 1}}} \right) = {2 \over {n + 1}} + {2 \over {{{\left( {n + 1} \right)}^2}}} + o\left( {{1 \over {{n^3}}}} \right)$$

$$\sqrt {n + 1} - \sqrt n \sim {1 \over {2\sqrt n }}$$

means we're interested in how

$${1 \over {2{n^{p/2}}}}{2 \over {n + 1}} + {1 \over {2{n^{p/2}}}}{2 \over {{{\left( {n + 1} \right)}^2}}} + {1 \over {2{n^{p/2}}}}o\left( {{1 \over {{n^3}}}} \right)$$

behaves for large $n$. Can you take it from there?

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Unfortunately, the series expansions were not introduced yet, so I don't really know what's going on here... –  trakos Dec 20 '12 at 18:41

Hint:

$$1)\quad \frac{1}{ (\sqrt{n+1}+\sqrt{n})^m }\sim \frac{1}{ (\sqrt{n})^m }=\frac{1}{n^{m/2}},$$

2) Use the integral test

$$\int_{2}^{\infty}\frac{\ln(x)}{x^a}dx$$ and find for what values of $a$ the integral converges.

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Sorry, the integrals were not yet introduced, so I don't know how to go from there. The ~ means something like "behaves the same way as"?. –  trakos Dec 20 '12 at 20:08
@trakos Ineed. More precisely, if $a_n\sim b_n$, then $\lim \frac{a_n}{b_n}=C$, $C$ a constant. –  Pedro Tamaroff Dec 22 '12 at 20:26
The integral test mentioned in 2) is not involved in the solution, in any way whatsoever. –  Did Dec 25 '12 at 11:22
@did:He has been asked to prove the convergence of the series. So, he can use integral test for this kind of series. Why do you think it has to be mentioned? –  Mhenni Benghorbal Dec 25 '12 at 12:18
@downvoter: What's the downvote for? It is an honesty issue. –  Mhenni Benghorbal Dec 25 '12 at 12:19