- #1

kwal0203

- 69

- 0

## Homework Statement

[tex] \sum_{x=2}^{\infty } \frac{1}{(lnx)^9} [/tex]

## Homework Equations

## The Attempt at a Solution

[tex] x \geqslant 2 [/tex]

[tex] 0 \leqslant lnx < x [/tex]

[tex] 0 < \frac{1}{x} < \frac{1}{lnx} [/tex]

From this we know that 1 / lnx diverges and I wanted to use this fact to show that 1 / [(lnx) ^ 9] diverges but at k >= 2 we have this:

[tex] 0 < \frac{1}{(lnx)^9} < \frac{1}{lnx} [/tex]

So it doesn't really work.

Is there a relation between 1/x and 1/[(lnx)^9] that I can use to solve the problem?

Any help would be appreciated, thanks!