Prove whether series converges or not? Does anyone know how to determine with proof whether the series
$$\sum_{n=1}^\infty\frac{1}{n^{2+\cos(2\pi\ln(n)) }}$$
converges?
 A: I am not completely sure in one argument below, but I think the series diverges. Let $\delta > 0$ be small and let $\varepsilon > 0$ be such that $\cos(\pi\pm\delta)=-1+\varepsilon$. Moreover, put $\tau := \delta/(2\pi)$. For $k\in\mathbb N$ define the set $\Delta_k^\tau := (e^{k+1/2-\tau},e^{k+1/2+\tau})$. Then we have $n\in\Delta_k^\tau$ if and only if $2+\cos(2\pi\ln(n))\in [1,1+\varepsilon)$ and the series is as least as large as
$$
\sum_k\sum_{n\in\Delta_k^\tau}\frac 1 {n^{1+\varepsilon}}.
$$
And now, I am not 100% sure. I claim that
$$
\sum_{n\in\Delta_k^\tau}\frac 1 {n^{1+\varepsilon}}\,\ge\,\int_{\Delta_k^\tau}\frac 1 {x^{1+\varepsilon}}\,dx.
$$
If not, then twice the left guy should do. Now,
$$
\int_{\Delta_k^\tau}\frac 1 {x^{1+\varepsilon}}\,dx = -\frac 1 {\varepsilon e^{\varepsilon/2}}\left(\frac 1{e^{\varepsilon\tau}} - \frac 1{e^{-\varepsilon\tau}}\right)\cdot e^{-\varepsilon k} = \frac 2 {\varepsilon e^{\varepsilon/2}}\sinh(\varepsilon\tau)e^{-\varepsilon k}.
$$
Summing over $k$, we get
$$
\frac 2 {\varepsilon e^{\varepsilon/2}}\sinh(\varepsilon\tau)\frac 1 {1-e^{-\varepsilon}} = \frac 1 \varepsilon \cdot\frac{\sinh(\varepsilon\tau)}{\sinh(\varepsilon/2)} = \frac 1 \pi\cdot\frac{\delta}{\varepsilon}\cdot\frac{\sinh(\varepsilon\tau)}{\varepsilon\tau}\cdot\frac{\varepsilon/2}{\sinh(\varepsilon/2)}.
$$
Now, we let $\delta\to 0$. Then, of course, also $\tau\to 0$ and $\varepsilon\to 0$. So, the last two factors tend to one. But $\delta/\varepsilon\to\infty$. Indeed, we have $\cos(\delta) = 1-\varepsilon$, so $\delta = \arccos(1-\varepsilon)$ and 
$$
\lim_{x\downarrow 0}\frac{\arccos(1-x)}{x} = \infty.
$$
This (hopefully) shows that the series diverges.
