# Converges or Diverges

• Nov 13th 2008, 12:01 PM
kl.twilleger
Converges or Diverges
$\displaystyle \sum_{k=2}^\infty \frac{4}{k ln k}$

should I turn the ln into an e? What rule should I use??
• Nov 13th 2008, 12:25 PM
Mathstud28
Quote:

Originally Posted by kl.twilleger
$\displaystyle \sum_{k=2}^\infty \frac{4}{k ln k}$

should I turn the ln into an e? What rule should I use??

Let $\displaystyle a_k=\frac{4}{k\ln(k)}$. Consider that $\displaystyle \forall{k}\in[2,\infty)~a_k>0\wedge{a_k}\in\mathcal{C}\wedge{a_k }\in\downarrow$ so the integral test applies. So this series shares convergence/divergence with $\displaystyle \int_2^{\infty}\frac{dx}{x\ln(x)}$
• Nov 13th 2008, 01:00 PM
albi
There is an alternative solution to this.

Theorem. If $\displaystyle a_1 > a_2 > ... > 0$ then series $\displaystyle \sum_{k = 1}^\infty a_k$ converges if and only if $\displaystyle \sum_{k = 1}^\infty 2^k a_{2^k}$ converges.
• Nov 13th 2008, 01:03 PM
Mathstud28
Quote:

Originally Posted by albi
There is an alternative solution to this.

Theorem. If $\displaystyle a_1 > a_2 > ... > 0$ then series $\displaystyle \sum_{k = 1}^\infty a_k$ converges if and only if $\displaystyle \sum_{k = 1}^\infty 2^k a_{2^k}$ converges.

Note to the poster, this is known as Cauchy's Condensation test and is rarely taught at a Calculus level course. It is useful in rare cases, but when it is applicable it is usually powerful.