Sum of alternating reciprocals of logarithm of 2,3,4... How to determine convergence/divergence of this sum?  
$$\sum_{n=2}^\infty \frac{(-1)^n}{\ln(n)}$$
Why cant we conclude that the sum $\sum_{k=2}^\infty (-1)^k\frac{k}{p_k}$, with $p_k$ the $k$-th prime, converges, since $p_k \sim k \cdot \ln(k)$ ?
 A: The Alternating Series Test, which is a special case of the Dirichlet Test, ensures the convergence of the first series.
To apply the Dirichlet test to $k/p_k$, one would have to show that the sequence $\{k/p_k\}$ has bounded variation.  That is,
$$
\sum_{k=1}^\infty\left|\frac{k}{p_k}-\frac{k+1}{p_{k+1}}\right|<\infty\tag{1}
$$
I don't know if $(1)$ is true.
A: You are correct.
The alternating series test suffices, no need to look at the dirichlet test.
Think about it, the sequence $|a|\rightarrow 0$ at $n\rightarrow\infty$, and decreases monotonically ($a_n>a_{n+1}$). That means that you add and remove terms that shrink to $0$. What you add you remove partially in the next term, and the terms shrink to 0. Eventually, your sum converges to a number but perhaps extremely slowly, but it will at least. If the terms didn't shrink to 0 but to a value $c$ or $a_{\infty}\rightarrow c$, then the series would be alternating around a value $middle\pm c$, hence not converge. We know this is not the case for $a_n=\frac{1}{p_n}$ (since there are infinitely many primes) and $a_n=\frac{1}{\log{n}}$, clearly $a_n$ tend to $0$ in both cases. For $a_n=\frac{n}{\log{n}}$, you are correct to assume that $\frac{p_n}{n}=\mathcal{O}\left(\log{n}\right)$. This means that $a_n=\frac{n}{p_n}=\mathcal{O}\left(\frac{1}{\log{n}}\right)\rightarrow 0$. From  "The kth prime is greater than k(log k + log log k−1) for k ≥ 2", $\frac{p_n}{n}=\frac{1}{a_n}$ is stuck between 
$\log n+\log\log n-1<\frac{p_n}{n}<\log{n}+\log\log n,$
then
$\frac{1}{\log n+\log\log n}<\frac{n}{p_n}<\frac{1}{\log{n}+\log\log n-1}.$
From the above inequality, we clearly see that the value $a_n=\frac{n}{p_n}$ is squished to $0$ by the bounds that vanish at infinity.
The sum accelerates so slowly that you may think that it alternates around a value $\pm c$, but this is not the case.
Similarly you can show that  $\sum_n \frac{(-1)^nn^{s}}{p_n }$ converges for $s\leq 1$, but will diverge for $Re(s)>1$.
And you have $\sum_n \frac{n^{s}}{p_n }$ converge if $Re(s)<0$. 
You see that modulating the sequence $a_n=\frac{n^{s}}{p_n }$ with $(-1)^n$ moves the region of convergence from $Re(s)<0$ to $Re(s)\leq 1$. The sum $\sum_n \frac{(-1)^nn^{s}}{p_n }$ is not differentiable for $
Re(s)=0$, therefore it should exhibit fractal like appearance on the imaginary line at the boundary. As it is for the Prime Zeta function, $
Re(s)=0$ is a natural boundary for the derivative of $\sum_n \frac{(-1)^nn^{s}}{p_n }$. For the sum itself, it is a "softer" natural boundary! 
