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I am able to prove divergence for $p<0$ or $p=0$.

How can I prove convergence/divergence for $p>0$.

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hint rewrite it as the power of e then divide numerator and denominator by n^p. –  Hardy Mar 23 '12 at 10:46
    
it seems interesting @Hardy right? –  dato datuashvili Mar 23 '12 at 10:47
    
it is an interesting expression and needs some thought, not sure if my earlier comment was right. –  Hardy Mar 23 '12 at 10:54
    
split the numerator as 1 + 2 + n^p so u get ln(1 + 1/(2 + n^p)) –  Hardy Mar 23 '12 at 10:56
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3 Answers

up vote 3 down vote accepted

Let $x_n=\log((3+n^p)/(2+n^p))$. This is $x_n=\log(1+y_n)$ with $y_n=1/(2+n^p)$.

If $p\lt0$, $y_n\to1/2$ and if $p=0$, $y_n\to1/3$. In both cases, $y_n\geqslant y^*$ for every $n$ with $y^*\gt0$ hence $x_n\geqslant\log(1+y^*)\gt0$ for every $n$ large enough and $\sum\limits_nx_n$ diverges.

If $p\gt0$, $y_n\to0$ hence $y_n/2\leqslant\log(1+y_n)\leqslant y_n$ for every $n$ large enough. Since $y_n\leqslant1/n^p$ for every $n$ and $y_n\geqslant1/(2n^p)$ for every $n$ large enough, this shows that the series $\sum\limits_nx_n$ behaves like $\sum\limits_n1/n^p$ hence it diverges for every $0\lt p\leqslant1$ and it converges for every $p\gt1$.

Finally, the series $\sum\limits_nx_n=\sum\limits_n\log((3+n^p)/(2+n^p))$ converges if and only if $p\gt1$.

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+1 for adding all the details. –  Hardy Mar 23 '12 at 10:59
    
@Didier Piau: This is rather off-topic, but would you mind having a look at this question of mine Divergent series expansion in Apéry's proof of the irrationality of $\zeta(2)$ and $\zeta(3)$ –  Américo Tavares Mar 25 '12 at 10:06
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You got: $$\ln \left( \frac{3+n^p}{2+n^p}\right) = \ln \left( 1+ \frac{1}{2+n^p}\right)$$ hence the well known asymptotics $\ln (1+y)\approx y$ as $y\to 0$ and $p>0$ yield: $$\ln \left( \frac{3+n^p}{2+n^p}\right) \approx \frac{1}{2+n^p} \approx \frac{1}{n^p}\; ;$$ therefore your series converges if and only if $p>1$.

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Hints:

$$ \log(1+x)=x+\mathcal{O}(x^2); \qquad \frac{3+n^p}{2+n^p}=1+\frac{1}{2+n^p} $$

Compare term-for-term with the usual $p$-series, $\sum n^{-p}$.

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nice one i was just getting to this conclusion too. –  Hardy Mar 23 '12 at 10:58
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