Is the series $\sum_{k=1}^{\infty}\frac{n_{k+1}-n_k}{n_{k+1}\log(n_{k+1})} $ convergent or not?

Question

For an arbitrary monotone subsequence $\{n_k\}\subset \mathbb{N}^+$ and $\lim_{k\to \infty}n_k = +\infty$. Does $$\sum_{k=1}^{\infty}\frac{n_{k+1}-n_k}{n_{k+1}\log(n_{k+1})} $$ convergent or not? Thanks very much.

Answer 1

This series may be convergent and may be not:

• L.F's comment gives a sequence $n_k=k$ where the series is divergent since in this case $$\frac{n_{k+1}-n_k}{n_{k+1}\log(n_{k+1})}=\frac{1}{(k+1)\log(k+1)}$$ and we show easily the divergence of the series by the integral test.
• By the sequence $n_k=\lfloor e^{k^2}\rfloor$ we have $$\frac{n_{k+1}-n_k}{n_{k+1}\log(n_{k+1})}\leq\frac{1}{\log(e^{(k+1)^2}-1)}\sim_\infty \frac{1}{k^2}$$ so it's a convergent series by comparison with the Riemann convergent series.