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

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?

3 Answers 3

1

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...2012-12-20
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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.

  • 0
    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"?.2012-12-20
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    @trakos Ineed. More precisely, if $a_n\sim b_n$, then $ \lim \frac{a_n}{b_n}=C$, $C$ a constant.2012-12-22
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    The integral test mentioned in 2) is not involved in the solution, in any way whatsoever.2012-12-25
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    @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?2012-12-25
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    @downvoter: What's the downvote for? It is an honesty issue.2012-12-25
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    As I said, the convergence of this series is NOT related to the convergence of the integral of $x\mapsto\log(x)/x^a$, for ANY value of $a$. (And I wonder what is the upvote for.)2012-12-25
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    @Did: Just read the problem carefully!!2013-02-07
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    I did, the indication in your post is wrong (and it seems you are falling back on your old habit of not listening to what people are explaining to you). There should be something vaguely related to $\log(x)$ when $x\to\infty$ in the original series, which would indicate its convergence: there is nothing of the sort.2013-02-07