Series...

**Also sprach Zarathustra** · May 14th 2010, 07:37 AM

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.

**chisigma** · May 14th 2010, 09:19 AM

Is...

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

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

Kind regards

$\chi$ $\sigma$

**Also sprach Zarathustra** · May 14th 2010, 10:27 AM

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...

**Plato** · May 14th 2010, 10:50 AM

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

**chisigma** · May 14th 2010, 10:58 AM

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

$\chi$ $\sigma$

**Also sprach Zarathustra** · May 14th 2010, 11:18 AM

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...

**Plato** · May 14th 2010, 11:47 AM

$\begin{array}{rcl} {\log (n^{\frac{1} {a}} )} & \leqslant & {n^{\frac{1} {a}} } \\ {\log (n)} & \leqslant & {an^{\frac{1} {a}} } \\ {\left[ {\log (n)} \right]^a } & \leqslant & {a^a n} \\ {\frac{1} {{\left[ {\log (n)} \right]^a }}} & \geqslant & {\frac{1} {{a^a n}}} \\ \end{array}$

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