# Absolute Convergence

1. Dec 2, 2007

### rocomath

$$\sum_{n=2}^{\infty}\frac{1}{(lnn)^{n}}$$

If I treat it as a geometric series, then when n=3, r is smaller than 1

$$\sum_{n=2}^{\infty}(\frac{1}{ln3})^3$$

What matters is that it's ultimately decreasing, would that be the correct approach? Even problem, no answer!

Thanks.

Last edited: Dec 2, 2007
2. Dec 2, 2007

What is r?

3. Dec 2, 2007

### rocomath

Well, as for an answer I would like to say that:

When n=3, r equals

$$|\frac{1}{ln3}|<1$$

Which is true, and so, what matters is that it is ultimately decreasing.

4. Dec 2, 2007

### EnumaElish

In the ratio test r = Limn-->infinity|Ln(n+1)-(n+1)/Ln(n)-n|.

If r < 1 then the series converges.

But ultimately the ratio test is about whether you have a decreasing sequence; so I think you can use the ratio test here.

Also, for n > 1, absolute convergence is the same as convergence because Ln(n) > 0.

Last edited: Dec 2, 2007
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