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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
up vote 2 down vote accepted

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.

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Using the fact that there are infinitely many prime pairs, we see that (1) diverges. Now, perhaps someone should disprove (1) without using facts whose proofs are unknown. – GEdgar Feb 16 '12 at 16:00
@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
@Sasha: insure and ensure are alternate spellings. There is more of a distinction in British English than in American English: "Our most recent evidence shows that the distinction between ensure and insure is made more often in British written English than in American written English, and a few commentators hold that insure is more common than ensure in American English." ("ensure, insure, assure," Merriam-Webster's Dictionary of English Usage, 1994) – robjohn Feb 16 '12 at 16:59
The sum over reciprocals of all primes in arithemtic progression of length k, converges for all k. – TROLLKILLER Feb 16 '12 at 17:03
@robjohn I see, I was not aware of this subtlety. Feel free to revert. – Sasha Feb 16 '12 at 17:11

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