Does this series converge or diverge?

Summation of 2/[k(lnk)^2]

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- May 6th 2010, 05:08 AMWartonMortonDoes this series converge or diverge?
Does this series converge or diverge?

Summation of 2/[k(lnk)^2] - May 6th 2010, 08:02 AMFailure
That's a typical stuation where you want to apply Cauchy's Condensation Test (Cauchy condensation test - Wikipedia, the free encyclopedia)

This theorem says that if the terms $\displaystyle a_k\geq 0$ are decreasing, the series $\displaystyle \sum_k^\infty a_k$ converges or diverges if and only if the series $\displaystyle \sum_n^\infty 2^n a_{2^n}$ converges or diverges, respectively (i.e. these two series behave the same way as regards convergence/divergence).

So let's take a closer look, therefore, at the series

$\displaystyle \sum_n^\infty 2^n \frac{1}{2^n \ln^2(2^n)}=\sum_n^\infty \frac{1}{n\ln^2(2)}=\frac{1}{\ln^2(2)}\sum_n^\inft y \frac{1}{n}$

which turns out to be essentially the same as the well known divergent series $\displaystyle \sum_n^\infty \frac{1}{n}$

Thus, the series $\displaystyle \sum_k^\infty \frac{1}{k\ln^2(k)}$**diverges**. (Blush)

Edited:

Plato is right, of course (he is almost always right): I made a silly mistake, since n comes out squared, one gets a series that converges:

$\displaystyle \sum_n^\infty 2^n \frac{1}{2^n \ln^2(2^n)}=\sum_n^\infty \frac{1}{n^2\ln^2(2)}=\frac{1}{\ln^2(2)}\sum_n^\in fty \frac{1}{n^2}$

So, since $\displaystyle \sum_n^\infty \frac{1}{n^2}$ converges, the original series**converges**. - May 6th 2010, 09:15 AMPlato