Does this series converge or diverge?

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October 29th 2009, 05:18 PM
uberbandgeek6

Does this series converge or diverge?

Does this series converge or diverge?

1 / ( (ln(n)) ^ (ln(n)) )

I'm not really sure how to approach this. It doesn't look like you can solve an indefinite integral, I can't find anything to used for the comparison test, and the ration test doesn't really simplify. What should I do? Thank you!
October 29th 2009, 07:50 PM
tonio

Quote:

Originally Posted by uberbandgeek6

Does this series converge or diverge?

1 / ( (ln(n)) ^ (ln(n)) )

I'm not really sure how to approach this. It doesn't look like you can solve an indefinite integral, I can't find anything to used for the comparison test, and the ration test doesn't really simplify. What should I do? Thank you!

You're forgetting, apparenty, the very nice Cauchy's Condensation Test: if a sequence $\{a_n\}$ is positive and monotonically descending to zero, the series $\sum\limits_{n=1}^\infty a_n$ converges iff the series $\sum\limits_{n=1}^\infty 2^na_{2^n}$ converges (and we can take any prime p instead of 2).
Apply this to your series and you'll find out it diverges (after, perhaps, you apply the n-th root test to the result of taking $2^na_{2^n}$ )

Tonio
October 30th 2009, 03:09 AM
uberbandgeek6

We never really learned that, so I would probably get docked points for not using a technique I already know.
October 30th 2009, 05:16 AM
tonio

Quote:

Originally Posted by uberbandgeek6

We never really learned that, so I would probably get docked points for not using a technique I already know.

Well, then first prove it. Try to google it or read any decent calculus book.

Tonio
October 30th 2009, 07:57 AM
Plato

Quote:

Originally Posted by uberbandgeek6

We never really learned that, so I would probably get docked points for not using a technique I already know.

That is a practice that many people don’t understand. But in many schools it is widely practiced. My guess is that your textbook & instructor want you to use the comparison test.

Here is the trick I learned thanks to Gillman’s text.
$\left(\ln(n)\right)^{\ln(n)}=n^{\ln[ln(n)]}$ .