# Convergence proof

• Mar 1st 2012, 03:07 PM
Dinghy
Convergence proof
The proposition is as follows:

If $\displaystyle a_n\geq0$ for all $\displaystyle n$ $\displaystyle \in \mathbb N$ and there is a $\displaystyle p>1$ such that $\displaystyle$\lim_{n\to\infty} n^pa_{n}$exists and is finite, then$\displaystyle \sum_{n=1}^\infty a_{n}$converges. I'm honestly not even sure what angle to come at this from, since I can't see what good knowing that limit exists does in trying to prove the convergence of some series. If someone could just give me a nudge in the right direction here I should be OK. • Mar 1st 2012, 03:37 PM Plato Re: Convergence proof Quote: Originally Posted by Dinghy The proposition is as follows: If$\displaystyle a_n\geq0 $for all$\displaystyle n\displaystyle \in \mathbb N$and there is a$\displaystyle p>1$such that$\displaystyle $\lim_{n\to\infty} n^pa_{n}$ exists and is finite, then $\displaystyle \sum_{n=1}^\infty a_{n}$ converges.

This is a straightforward application of the limit comparison test.

Note that if $\displaystyle p>1$ then $\displaystyle \sum\limits_n {\frac{1}{{n^p }}}$ converges.
• Mar 1st 2012, 03:50 PM
Dinghy
Re: Convergence proof
Quote:

Originally Posted by Plato
This is a straightforward application of the limit comparison test.

Note that if $\displaystyle p>0$ then $\displaystyle \sum\limits_n {\frac{1}{{n^p }}}$ converges.

Legitimately embarrassed about not having seen that. This problem had been talked up so much that it didn't even occur to me that it could be such a straightforward application of one of the simpler convergence tests. One thing though, when you say $\displaystyle p>0$, you mean $\displaystyle p>1$, right?

Thanks a lot.
• Mar 1st 2012, 04:11 PM
Plato
Re: Convergence proof
Quote:

Originally Posted by Dinghy
you mean $\displaystyle p>1$, right?

Yes, of course. Corrected!
• Mar 1st 2012, 04:26 PM
Dinghy
Re: Convergence proof
Actually I just noticed, I don't think you can apply the limit comparison test, as technically $\displaystyle \sum_{n=1}^\infty a_{n}$ isn't necessarily a positive series. The proposition says that $\displaystyle a_{n}\geq 0$, whereas a positive series requires that $\displaystyle a_{n}>0$, so I think I'm back at square one.

edit: but I guess the Limit Comparison test doesn't strictly require it to be a positive series, just that $\displaystyle a_{n}\geq 0$...

Can't shake the feeling that my math professor is messing with me with the "$\displaystyle \sum_{n=1}^\infty a_{n}$ isn't necessarily a positive series!!" warning.
• Mar 1st 2012, 04:44 PM
Plato
Re: Convergence proof
Quote:

Originally Posted by Dinghy
Actually I just noticed, I don't think you can apply the limit comparison test, as technically $\displaystyle \sum_{n=1}^\infty a_{n}$ isn't necessarily a positive series. The proposition says that $\displaystyle a_{n}\geq 0$, whereas a positive series requires that $\displaystyle a_{n}>0$, so I think I'm back at square one.k

edit: but I guess the Limit Comparison test doesn't strictly require it to be a positive series, just that $\displaystyle a_{n}\geq 0$...

Can't shake the feeling that my math professor is messing with me with the "$\displaystyle \sum_{n=1}^\infty a_{n}$ isn't necessarily a positive series!!" warning.

I don't understand what you are concerned about.
Look at the link I provided.
It requires that $\displaystyle a_n\ge 0~\&~b_n\ge 0~.$
• Mar 1st 2012, 04:48 PM
Dinghy
Re: Convergence proof
Yeah, sorry, my Calc 2 professor defined the LCT really lazily (said that it only applies to positive series), and once I get something in my head it's really hard to get it out.