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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?

  • 0
    Unfortunately, the series expansions were not introduced yet, so I don't really know what's going on here...2012-12-20
0

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
    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