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