if a>1 then the series $\sum\limits_{k = 1 }^ \infty \frac{ 1}{k^a} $ converges
What is the limit value of $a$ for the series below?
$\sum\limits_{k = 1 }^ \infty \frac{ \ln(k)}{k^a} $
My steps to find the value: if $x > 1$ then $\ln x
thus
$\sum\limits_{k = 1 }^ \infty \frac{ \ln(k)}{k^a} < \sum\limits_{k = 1 }^ \infty \frac{ k}{k^a} = \sum\limits_{k = 1 }^ \infty \frac{ 1 }{k^{a-1}}$
if $a>2$, It is sure that $\sum\limits_{k = 1 }^ \infty \frac{ \ln(k)}{k^a}$ the series converges. But we can find a lower value for the series.
Could you please help me to find the exact value of $a$ that the series converge?
Thanks for answers