My instinct is to use the root test, and try to show thatQuote:

Prove that

converges.

.

But unfortunately I don't know how to show this either.

Are there any tricks I can use on these nth root proofs which might help here?

Thanks!

Printable View

- Sep 10th 2009, 06:49 PMhatsoff[SOLVED] prove that this series to the nth^2 power convergesQuote:

Prove that

converges.

.

But unfortunately I don't know how to show this either.

Are there any tricks I can use on these nth root proofs which might help here?

Thanks! - Sep 10th 2009, 07:00 PMKrizalid
Note that for each is and then besides the series converges by the integral test, thus the original series does converge.

----

Of course your idea does work too. In order to prove that as we have

so it's easy to see that as and hence, the result. - Sep 10th 2009, 08:04 PMhatsoff
Thank you very much, but I could still benefit from further assistance.

Once we have , the proof is easy for me. However, I am uncertain how you arrived at that inequality.

We have by definition

.

I have no idea how to prove that this series is increasing, but let's say for a moment that I could. In that case, we have, for each ,

.

Then algebraically we can show that

flipping:

.

Substitute back to get

,

and expand:

.

But this isn't good enough to get to our goal, which, again, is:

.

Apparently I need to learn more about exponents. Any more tricks would be much appreciated.