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

A: Using Abel's summation (integration by parts):
$$\begin{split}
\sum_{k=2}^n \dfrac{1}{k^{\epsilon}}\left(\dfrac{1}{\log k}-\dfrac{1}{\log(k+1)}\right) &= \left(\sum_{k=2}^n \dfrac{1}{k^{\epsilon}}\dfrac{1}{\log k}\right)-\left(\sum_{k=3}^{n+1}\dfrac{1}{(k-1)^{\epsilon}}\dfrac{1}{\log k}\right)\\
&=\frac 1 {2^\varepsilon}\frac 1 {\log 2}-\frac 1 {n^\varepsilon}\frac 1 {\log (n+1)}+\sum_{k=3}^n \left( \frac 1 {k^\varepsilon} - \frac 1 {(k-1)^\varepsilon}\right)\frac 1 {\log k}
\end{split}$$
Also $$\begin{split}
\frac 1 {k^\varepsilon} - \frac 1 {(k-1)^\varepsilon} &= \frac 1 {k^\varepsilon} \left( 1 - \frac 1 {\left ( 1 - \frac 1 k\right)^\varepsilon}\right)\\
&= \frac 1 {k^\varepsilon} \left( -\frac \varepsilon k + \mathcal O\left(\frac 1 {k^2}\right)  \right)\\
&= -\frac \varepsilon {k^{\varepsilon+1}} + \mathcal O\left(\frac 1 {k^{\varepsilon+2}}\right)  \\
\end{split}$$
Thus the series converges.
To your second question, if you replace $\log(k)$ with $f(k)$, you see that the convergence or divergence of the original series depends on that of $$\sum \frac 1 {k^{\varepsilon+1}} \frac 1 {f(k)}$$
and since you assume $f$ to be non-decreasing, the term of the series is bounded from above by $\frac 1 {k^{\varepsilon+1}} \frac 1 {f(1)}$, and thus the series always converges.
A: Perhaps all of us have just thinking too much:
\begin{align*}
\sum_{k\geq n_{0}}\dfrac{1}{k^{\epsilon}}\left(\dfrac{1}{f(k)}-\dfrac{1}{f(k+1)}\right)&\leq\sum_{k\geq n_{0}}\left(\dfrac{1}{f(k)}-\dfrac{1}{f(k+1)}\right)\\
&=\dfrac{1}{f(n_{0})}-\lim\dfrac{1}{f(n)}\\
&=\dfrac{1}{f(n_{0})}-\dfrac{1}{\sup f(n)}.
\end{align*}
