Thread: Series

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

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

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

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

Thus the limit exist when $\displaystyle p + 1 < 0$ 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

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

Thanks

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

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 :

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

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

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If you want to show it's decreasing :

x(1+x²)^p

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

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

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

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


etc.