I need to check if the next infinite series convergent:
Sigma (k=2-->inf) 1/{ln(k)}^a (when a real positive number)
Thank you.
The expression $\displaystyle \lim_{k \rightarrow \infty} \frac{\ln^{a} k}{k} = 0$ means that exists a $\displaystyle K$ so that for any $\displaystyle k>K$ is $\displaystyle \frac{1}{\ln^{a} k} > \frac{1}{k}$. Now the series $\displaystyle \sum_{k=2}^{\infty} \frac{1}{k}$ diverges so that the series $\displaystyle \sum_{k=2}^{\infty} \frac{1}{\ln^{a} k}$ also diverges...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
$\displaystyle \begin{array}{rcl}
{\log (n^{\frac{1}
{a}} )} & \leqslant & {n^{\frac{1}
{a}} } \\
{\log (n)} & \leqslant & {an^{\frac{1}
{a}} } \\
{\left[ {\log (n)} \right]^a } & \leqslant & {a^a n} \\
{\frac{1}
{{\left[ {\log (n)} \right]^a }}} & \geqslant & {\frac{1}
{{a^a n}}} \\
\end{array} $