# Does this tricky series converge?

$$\sum_2 \frac{\cos(\log{n})}{n\log{n}}$$

The naive attempt is to use dirichlet's test to falsely claim that $\cos(\log{n})$ has bounded partial sums, but I don't think it works.

I am also trying a difference of sum and integral type of strategy but am not sure where to go from it.

Finally, as A.S pointed out below in the comments, the integral test does not apply, since cos(logn) does not behave monotonically...

Any ideas are welcome.

Thanks,

• @Gregory Because intergral test doesn't apply to unmonotonic sequences. – A.S. Dec 25 '15 at 1:00
• @User001 He's suggesting grouping all consecutive terms with the same sign. It's a good idea but I think the argument is incomplete, it's not clear to me that the terms once grouped do go to zero, it's not enough that cosine is bounded because the number of consecutive terms with the same sign goes to infinity (I think). – Gregory Grant Dec 25 '15 at 1:23
• No reindexing. Just decomposing the sum into pos/neg terms lumped together which signify "maximal deviations". @Gregory are you missing the second part of the argument the sum of 1/denom over the $k$-th range goes down as $\Theta(1/k)$? – A.S. Dec 25 '15 at 1:23
• No. For example, $\log{2k}-\log{k}=\log 2<\infty$ even though $2k-k\to\infty$. See this for big $\Theta$ notation. – A.S. Dec 25 '15 at 1:58
• @User001 Thank you it's very nice of you to post these thank yous, you are a gentleman (or woman, ambiguous user name) and a scholar. – Gregory Grant Dec 26 '15 at 6:58

First examine the corresponding integral:

$$\int_2^{\infty} \frac{\cos(\log(x))}{x \log(x)}dx = \int_{log(2)}^{\infty} \frac{\cos(u)}{u} du = \left[ \frac{\sin(u)}{u} \right]_{log(2)}^{\infty} + \int_{log(2)}^{\infty} \frac{\sin(u)}{u^2}du$$ this is convergent.

Next we look at the derivative of the function \begin{align}f(x) &= \frac{\cos(\log(x))}{x \log(x)} \\ f'(x) &= \frac{-1/x \cdot\sin(\log(x))\cdot x \log(x) + \cos(\log(x))(\log(x)+1)}{(x \log(x))^2}\\ |f'(x)| &< \frac{2}{x^2}\end{align} so we can compare the terms of sum and the integral over intervals of length 1: \begin{align} \left| \frac{\cos(\log(k))}{k \log(k)} - \int_k^{k+1} \frac{\cos(\log(x))}{x \log(x)} dx \right| &\leq \int_k^{k+1} | f(k) - f(x) | dx \\ &\leq \max_{k<x<k+1} | f'(x) | < \frac{2}{k^2} \end{align}

Since the integral and the sum of the differences both are convergent, the original series is convergent.

• I think that some text explaining why the things you mention are relevant would be a good addition. – Aloizio Macedo Dec 25 '15 at 1:23
• difference $f(x) - f(k) = (x-k)f'(\xi)$ – mlu Dec 25 '15 at 1:45
• I hope its getting clearer – mlu Dec 25 '15 at 1:57
• Conditionally convergent, yes. – mlu Dec 25 '15 at 9:29
• The important part is that we look at cos in widely spaced points log(k) to make the derivative decrease like $1/x^2$. The derivative of $\cos(x)/x$ only decreases like $1/x$ – mlu Dec 25 '15 at 15:32