# Convergence (series)

• May 26th 2012, 09:06 AM
fqqs
Convergence (series)
I have to assess if $\displaystyle \sum_{n=1}^{ \infty } \frac{1}{1+ n^{2} \cdot x^{2} }$ is convergent (uniformly convergent)

• May 26th 2012, 10:08 AM
Plato
Re: Convergence (series)
Quote:

Originally Posted by fqqs
I have to assess if $\displaystyle \sum_{n=1}^{ \infty } \frac{1}{1+ n^{2} \cdot x^{2} }$ is convergent (uniformly convergent)

Think of the limit comparison test.

What can you say about $\displaystyle \left( {\frac{{{n^2}}}{{1 + {n^2}{x^2}}}} \right) \to~?$
• May 26th 2012, 10:20 AM
fqqs
Re: Convergence (series)
Quote:

Originally Posted by Plato
Think of the limit comparison test.

What can you say about $\displaystyle \left( {\frac{{{n^2}}}{{1 + {n^2}{x^2}}}} \right) \to~?$

It goes to $\displaystyle \frac{1}{x^{2}}$?
• May 26th 2012, 10:48 AM
Plato
Re: Convergence (series)
Quote:

Originally Posted by fqqs
It goes to $\displaystyle \frac{1}{x^{2}}$?

What does that tell you about the values of $\displaystyle x$ for which it converges or diverges?
• May 26th 2012, 11:05 AM
fqqs
Re: Convergence (series)
it converges for every $\displaystyle x \ne 0$

?
• May 26th 2012, 11:48 AM
Plato
Re: Convergence (series)
Quote:

Originally Posted by fqqs
it converges for every $\displaystyle x \ne 0$

Correct
• May 26th 2012, 12:18 PM
fqqs
Re: Convergence (series)
but that only means that it is pointwisely convergent for every $\displaystyle x \ne 0$, yes?

so i need to check if it is pointwisely convergent somewhere , yes?

maybe I should use cases when $\displaystyle |x| < 1$ and when $\displaystyle |x| \ge 1$ and use Weierstrass?
• Jun 1st 2012, 10:33 AM
fqqs
Re: Convergence (series)
upppp