5
$\begingroup$

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

  • 1
    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 $b_n=\frac1{\sqrt{n}}+\frac{(-1)^n}n$.2012-02-16

2 Answers 2

2

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.

  • 0
    @Sasha: both are fine, and since "ensure" is better in Britain, there is no reason to revert. It was educational to look things up :-)2012-02-16