Series

**11rdc11** · Sep 3rd 2008, 12:17 PM

Ok so I'm working on this problem but have problem understanding it. I got the right answer but I made an assumption to get there. I need help proving that assumption.

Here is the problem

Find the values of p for which the series in convergent

$\sum_{n=1}^{\infty} n(1+n^2)^p$

Ok so I used the integral test because f(x) = $x(1+ x^2)^p$ is continous, positive, and decreasing(this is where I made the assumption that it was decreasing).

$\int_{1}^{\infty} x(1+x^2)^{p}dx = \lim_{t \to \infty}\bigg[\frac{(1+x^2)^{p+1}}{2p+2}\bigg]_1^t = \lim_{t \to \infty}\bigg(\frac{1}{2}\bigg)\bigg(\frac{(1+t^2)^ {p+1}}{p+1}\bigg) - \frac{2^p}{p+1}$

Thus the limit exist when $p + 1 < 0$ so the series converges when p< -1.

Is that right? Also how do I show that the function is decreasing(the assumption I need to prove). I took the 1st derivative and equalled it to less than 0 to show me where it was decreasing but ran into problems.

This is what I get stuck on with the derivative

$x^2(1+2p)< -1$

Thanks

**11rdc11** · Sep 3rd 2008, 01:50 PM

Can I say the function is decreasing since when p is negative the denominator is growing faster than the numerator so it is decreasing?

**Moo** · Sep 4th 2008, 12:44 AM

Hello,

Yes, it is continuous.

Actually, my reasoning is as follows :
The series {a_n} converges ---> lim a_n = 0

By taking the contrapositive of this assumption, it's evident to say that lim a_n has to be 0.

Assuming p<0 (otherwise, one can see it is not possible), let q=-p for more convenience :

$\frac{n}{(1+n^2)^q} \to_{\substack{n \to \infty}} \frac{n}{n^{2q}}$

---> for it to tend to 0, q has to be > 1/2, which is p<-1/2. Necessary condition, not sufficient.

-----------------------------------------------
If you want to show it's decreasing :

x(1+x²)^p

Its derivative is $(1+x^2)^p+p 2x^2(1+x^2)^{p-1}=(1+x^2)^{p-1} (1+x^2+2px^2)$

$(1+x^2)^{p-1} \ge 0$ because $x \ge 1$

$1+(2p+1)x^2$
We want to find p such that it is $\le 0$

if $2p+1\ge 0$ ---> (2p+1)x²+1 is always positive. Which is not possible.
if $2p+1<0$ , since x²>1, (2p+1) has to be < -1, that is to say p<-1

etc.

Thread: Series

LinkBack

Thread Tools

Display

Series

Similar Math Help Forum Discussions

Show that a series converges absolutely given convergence of two related series

Exploiting geometric series with power series to find Taylors series

question about power series, taylor series, maclaurin series

Does the subtraction of 2 diverging Series mean the resulting Series is convergent?

calculus II areas about axis, pressure, infinite series, MacLaurin series

Search Tags