# How to determine the convergence of this series.

I have tried several methods to try to determine the convergence of the following series, to no avail:

$$\sum_{n=2}^{\infty}\frac{1}{n\cdot \log^{2+\epsilon}{n}} : \forall\epsilon>0$$

First thing I do is consider this other series:

$$\sum_{n=2}^{\infty}\frac{1}{n\cdot \log^2{n}}$$

Which if converges, will imply the previous one converges as well as it's smaller. The problem is I can't find the convergence of this one. I've tried the ratio and root tests but they are both inconclusive. What am I missing?

EDIT: Forgot to mention that I'm not supossed to be able to use the Integral test yet, sorry for the inconvenience.

• Compare the series to the integral. That immediately yields the convergence. Jan 2 '14 at 0:08
• Oops sorry, forgot to mention that I'm not supossed to be able to use the Integral test yet, sorry for the inconvenience. Jan 2 '14 at 0:08
• $\displaystyle{\large\sum_{n = 2}^{\infty} {1 \over n\log^{2}\left(n\right) } = 2.10974}$ Jan 2 '14 at 2:59

If you can't use the integral comparison, use the Cauchy condensation test. Since the terms are monotonically decreasing and positive,

$$\sum_{n=2}^\infty a_n < \infty \iff \sum_{k=1}^\infty 2^k\cdot a_{2^k} < \infty.$$

Here,

$$\sum_{k=1}^\infty \frac{2^k}{2^k\cdot (\log (2^k))^2} = \sum_{k=1}^\infty \frac{1}{(\log 2)^2 k^2} < \infty.$$

• Thanks! And to think that I initially discarded the Cauchy condensation test because I "didn't feel" it was going to be successful. Jan 2 '14 at 0:17
• @LMartin i think this tests is the most useful one out of the standard ones taught at the beginning of uni. The other is probably dirichlet. Jan 2 '14 at 0:25
• Oh tha'ts a good advice, I'll keep that in mind for future problems. Jan 2 '14 at 0:31

Since the integral $$\displaystyle\int_2^\infty\frac{dx}{x\log^{2+\varepsilon}x}=\displaystyle\int_{\log 2}^\infty\frac{du}{u^{2+\varepsilon}}$$ converges for any $\varepsilon\geq 0$, you may apply the integrating test as well.