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Which values for $a$ will make series convergent or divergent?

$$ \sum_{k=1}^{\infty}\frac{1}{(2k+1)\ln^a(2k+1)} $$

I have been following this tutorial But now I'm stuck. I haven't seen examples as this one above. There is additional parameter $a$ which values I have to find myself. Could anyone suggest me how should I work this time then? Maybe I just choose randomly some values for $a$? Like $a=-2$, $0$ and $2$?

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up vote 4 down vote accepted

This nature of this series is ruled by the so-called Bertrand criterion, which tells you that $\displaystyle \sum_{n \geqslant 2} \frac{1}{n^{\alpha} \ln^{\beta}(n)}$ converges if and only if $\alpha >1$ or $\alpha=1$ and $\beta >1$.

More precisely, in your case, the general term is positive, and equivalent to $\displaystyle\frac{\rm cte}{n \, \ln^a(n)}$, so the series converges iff $a>1$.

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hey, thank you. i'll look further now for Bertrand criterion – Skyzer May 15 '11 at 14:52

If you are comfortably familiar with the tutorial series that you have been studying from, look at their discussion under integral test.

By following what they say word for word, you can see that the given series converges precisely if $$\int_{x=1}^\infty \frac{1}{(2x+1)\ln^{a}(2x+1)}dx$$ converges.

The integral is easy to deal with. To find an antiderivative, just make the substitution $u=\ln(2x+1)$. Then $du=2dx/(2x+1)$, so for the antiderivative you want to find $$\int \frac{du}{2u^a}$$ and the definite integral is $$\int_{u=\ln 3}^\infty \frac{du}{2u^a}$$

I imagine that you are familiar with the fact that this "improper" integral diverges if $a \le 1$ and converges if $a>1$. If you are not, just calculate the integral from $\ln 3$ to $M$, and examine what happens as $M\to\infty$. This is quite easy, since you know very well how to integrate $u^{-a}$.

We conclude that the original series converges if $a>1$ and diverges otherwise.

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