 # For what values for m does <munderover> <mo movablelimits="false">&#x2211;<!-- ∑ --> <mr Michelle Mendoza 2022-07-03 Answered
For what values for m does
$\sum _{k=2}^{\mathrm{\infty }}\frac{1}{\left(\mathrm{ln}k{\right)}^{m}}$
converge?
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For your first series
Let $\alpha ,\beta \in \mathbb{R}$
$\sum _{k=2}^{\mathrm{\infty }}\frac{1}{{k}^{\alpha }\mathrm{ln}\left(k{\right)}^{\beta }}$
is convergent iif $\left(\alpha >1\right)$ or
So in your case it is not convergent. As
$\frac{\sqrt{k}}{\mathrm{ln}\left(k{\right)}^{\beta }}\to \mathrm{\infty }$
we can find ${k}_{0}$ such that for all $k\ge {k}_{0}$
$\frac{\sqrt{k}}{\mathrm{ln}\left(k{\right)}^{\beta }}>1$
which yields
$\frac{1}{\mathrm{ln}\left(k{\right)}^{\beta }}>\frac{1}{\sqrt{k}}$
The right-hand side behind divergent, your series is divergent. These series are called Bertrand series.