# Thread: Series...

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...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by chisigma
Is...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
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 $\displaystyle a>0~\&~n\ge 3$
$\displaystyle \frac{1}{[log(n)]^a}\ge \frac{1}{a^a\cdot n}$.

5. 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

$\displaystyle \chi$ $\displaystyle \sigma$

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

For Plato, how can I prove this?

Thank you...

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