# 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 $p_k \sim k \cdot \ln(k)$ ?

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Alternating series test (by the way, the sum should start at $n=2$ to avoid division by 0). – David Mitra Feb 16 '12 at 15:25
maybe you should start at $n=2$. – draks ... Feb 16 '12 at 15:25
Provided the summation begins at $n=2$, this is an alternating series hence the usual test gives the answer. – Did Feb 16 '12 at 15:26
Re your second question, added later on: this is because the alternating series test requires the unsigned sequence to be decreasing. Even when $a_n\gt0$, $b_n\gt0$ and $a_n/b_n\to1$, $\sum(-1)^na_n$ and $\sum(-1)^nb_n$ may behave differently. Example: $a_n=\frac1{\sqrt{n}}$ and $b_n=\frac1{\sqrt{n}}+\frac{(-1)^n}n$. – Did Feb 16 '12 at 16:06

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.
@GEdgar: the contribution to $(1)$ from a prime pair would be $$\left|\frac{k}{p_k}-\frac{k+1}{p_k+2}\right| = \left|\frac{2k-p_k}{p_k(p_k+2)}\right|\sim\frac{1}{p_k}$$ However, this is only contributed over prime pairs. Is it known that the sum of the reciprocals of prime pairs converges or diverges? – robjohn Feb 16 '12 at 16:35