# Convergence of a series

• November 26th 2008, 12:31 PM
davidmccormick
Convergence of a series
Consider

$f (x) = \sum_{n=1} ^ {\infty} 1/{n(1 + nx^2)}$

(a) For what values of $x$ does the series converge?
(b) On what intervals of the form $(a,b)$ does the series converge uniformly?
(c) On what intervals of the form $(a,b)$ does the series fail to converge uniformly?
(d) Is $f$ continuous at all points where the series converges?

I would appreciate any help. Meanwhile I am trying to do the question by myself and would let everyone know of any breakthroughs.

Thanks
• November 26th 2008, 12:51 PM
davidmccormick
Ok, I have done part (a). Using the limit comparison test with the series $1/n^2$ which we know converges we compute the limit to be $x^2$ which in reals is always positive provided it's non zero. If $x = 0$ well then we get the harmonic series which everyone knows diverges.(Nod)
• November 26th 2008, 12:57 PM
Mathstud28
Quote:

Originally Posted by davidmccormick
Consider

$f (x) = \sum_{n=1} ^ {\infty} 1/{n(1 + nx^2)}$

(a) For what values of $x$ does the series converge?
(b) On what intervals of the form $(a,b)$ does the series converge uniformly?
(c) On what intervals of the form $(a,b)$ does the series fail to converge uniformly?
(d) Is $f$ continuous at all points where the series converges?

I would appreciate any help. Meanwhile I am trying to do the question by myself and would let everyone know of any breakthroughs.

Thanks

Is this $\sum_{n=1}^{\infty}\frac{1+nx^2}{n}$ or $\sum_{n=1}^{\infty}\frac{1}{n(1+nx^2)}$. I think it is the latter.

a) All except 0 consider that $\lim_{n\to\infty}\frac{\frac{1}{n(1+nx^2)}}{\frac{ 1}{n(n+1)}}=x^2$

b) Use the Weirstrass M-test. For which thing to compare it to consider what interval you may definitely say that $\left|\frac{1}{n(1+nx^2)}\right|\leqslant\frac{1}{ n(n+1)}$ and then consider the other interval serperately

c) This should be found during b0

d) Use the fact that if $\sum_{n=0}^{\infty}f_n(x)$ is uniformly convergent on $[a,b]$ and $f_n(x)\in\mathcal{C}$ than so is $f(x)=\sum_{n=0}^{\infty}f_n(x)$
• November 26th 2008, 01:11 PM
davidmccormick
It is indeed the latter. Thank you for your help.
• November 27th 2008, 04:41 AM
davidmccormick
Parts (a) and (d) are fine. For part (b) I managed to show that the series converges uniformly on the intervals $(-{\infty},-1]$ and $[1, {\infty})$ but have no idea how to show whether or not uniform convergence applies to (-1,1) interval. Can you explain a little more please? thanks.
• November 27th 2008, 09:09 AM
Mathstud28
Quote:

Originally Posted by davidmccormick
Parts (a) and (d) are fine. For part (b) I managed to show that the series converges uniformly on the intervals $(-{\infty},-1]$ and $[1, {\infty})$ but have no idea how to show whether or not uniform convergence applies to (-1,1) interval. Can you explain a little more please? thanks.

Try considering the intervals $\left(a\ne{0},1\right)$ and $\left(-1,a\ne{0}\right))$ and that $\left|\frac{1}{n
(1+nx^2)}\right|\leqslant\frac{1}{n(1+\min_{(a\ne{ 0},1)}\left\{x^2\right\}n)}$
and use the Limit comparison test on the right hand series.
• November 27th 2008, 10:30 AM
davidmccormick
What series can we compare the series on the right hand side of the inequality to (in the limit comparison test)? Also, if we can show this it means that our original series is uniform convergent everywhere except for
$x = 0$, and so the answer to part (c) is any interval which doesn't contain 0, right?
• November 27th 2008, 11:59 AM
Opalg
Quote:

Originally Posted by davidmccormick
What series can we compare the series on the right hand side of the inequality to (in the limit comparison test)?

If $0 then $0\leqslant\frac1{n(1+nx^2)}\leqslant\frac1{n(1+na^ 2)}\leqslant \frac1{n^2a^2}$ and you can use the M-test. Similarly on the interval $-1\leqslant x\leqslant -a$.

Quote:

Originally Posted by davidmccormick
so the answer to part (c) is any interval which doesn't contain 0, right?

Any closed interval which doesn't contain 0.
• November 27th 2008, 02:34 PM
davidmccormick
thanks very much for your help.