1. ## Series...

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.

2. Is...

$\lim_{k \rightarrow \infty} \frac{\ln ^{a} k} {k}= 0$

... no matter which is $a$, so that...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
Is...

$\lim_{k \rightarrow \infty} \frac{\ln ^{a} k} {k}= 0$

... no matter which is $a$, so that...

Kind regards

$\chi$ $\sigma$
This is not helping me, I can't use the limit test:
lim(n-->inf) an/bn =L when 0<L<inf thet u can say something about the series...

4. Use simple comparsion. For all $a>0~\&~n\ge 3$
$\frac{1}{[log(n)]^a}\ge \frac{1}{a^a\cdot n}$.

5. The expression $\lim_{k \rightarrow \infty} \frac{\ln^{a} k}{k} = 0$ means that exists a $K$ so that for any $k>K$ is $\frac{1}{\ln^{a} k} > \frac{1}{k}$. Now the series $\sum_{k=2}^{\infty} \frac{1}{k}$ diverges so that the series $\sum_{k=2}^{\infty} \frac{1}{\ln^{a} k}$ also diverges...

Kind regards

$\chi$ $\sigma$

6. Originally Posted by Plato
Use simple comparsion. For all $a>0~\&~n\ge 3$
$\frac{1}{[log(n)]^a}\ge \frac{1}{a^a\cdot n}$.

For Plato, how can I prove this?

Thank you...

7. $\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}$