On the Convergence of Series Will Jagy has provided the answer for the following question, the series is convergent:
I come across with the following question about the convergence of series:
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\left(\dfrac{1}{\log k}-\dfrac{1}{\log(k+1)}\right),
\end{align*}
where $\epsilon\in(0,1)$ is fixed.
The only thing I can do is no more than
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\dfrac{1}{(\log k)(\log(k+1))}\log(1+1/k).
\end{align*}
I propose the series is actually divergent. The term $(\log k)(\log(k+1))$ is slower than any $k^{\eta}$ for $\eta\in(0,1)$, on one hand the series has the lower bound
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\dfrac{1}{k^{\eta}}\log(1+1/k),
\end{align*}
which we can put $\epsilon+\eta\leq 1$. But I cannot put a lower bound of the rate of $\log(1+1/k)$, any idea?
EDIT:
Stefan Lafon has answered the following question, the answer is negative, the series will always be convergent:
So I am looking for a nonnegative increasing function $f$ such that
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\left(\dfrac{1}{f(k)}-\dfrac{1}{f(k+1)}\right)
\end{align*}
is divergent, but I cannot find one. Initially I was thinking the $\log$ thing will do, but Will Jagy disproves that. So what is such an $f$ for instance?

For $0 < t < 1,$
$$ t - \frac{t^2}{2} < \log (1+t) < t  $$
$$ \frac{1}{k} - \frac{1}{2k^2} < \log \left(1+ \frac{1}{k} \right) <  \frac{1}{k} $$
$$  \frac{2k-1}{2k^2} < \log \left(1+ \frac{1}{k} \right) <  \frac{1}{k} $$
$$  \frac{k}{2k^2} < \frac{2k-1}{2k^2} < \log \left(1+ \frac{1}{k} \right) <  \frac{1}{k} $$
$$  \frac{1}{2k}  < \log \left(1+ \frac{1}{k} \right) <  \frac{1}{k} $$

