Find the sum of series (n^2 / (n+!)) x^2 ?

Question

anonymous · Answer

$$\sum_{n=1}^{Infinity} \frac {n^2} {n+1} x^2$$

anonymous · Answer

$$\sum_{n=1}^\infty\frac{n^2}{n+1}x^2$$Well, observe:$$\frac{n^2}{n+1}=\frac{n^2-1+1}{n+1}=\frac{n^2-1}{n+1}+\frac1{n+1}=n+1+\frac1{n+1}$$Since $n\to\infty$ this sum cannot converge...

anonymous · Answer

It converges for abs(x) < 1

anonymous · Answer

http://www.wolframalpha.com/input/?i=sum_%7Bn%3D0%7D%5Einfty+n%5E2%2F%28n%2B1%29+x%5E2

anonymous · Answer

Wolfram is not always correct, for example take x = 0.

anonymous · Answer

Obviously for $x=0$ the entire summation vanishes... but it does not converge for $x\ne0$... I thought you would be able to piece that together yourself...

anonymous · Answer

$x^2$ is invariant and does not change in the summation because it does not depend on $n$  -- thus we can pull it out:$$x^2\sum_{n=0}^\infty\left(1+n+\frac1{n+1}\right)$$

anonymous · Answer

oops, $n-1+\frac1{n+1}$ is our summand -- messed up the sign on $1$

anonymous · Answer

But I've found the area of convergence using Dirichlet formula $$\lim_{n \rightarrow \infty} \left|  \frac {a_{n+1}} {a_{n}} \right|=1$$ which means that it should converge for $$\left| x \right|<1$$ since it diverges for x = 1 and x = -1 (and of course $$\left| x \right| > 1$$)

anonymous · Answer

Uh... what you've used is known as the ratio test: http://en.wikipedia.org/wiki/Ratio_test as you see, $L=1$ is inconclusive

jhannybean · Answer

Oh. The Ratio Test.

anonymous · Answer

Consider the harmonic series $\sum\limits_{n=1}^\infty\frac1n$:$$\lim_{n\to\infty}\left|\frac{n}{n+1}\right|=\lim_{n\to\infty}\frac{n}{n+1}=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)=1$$yet we already know said series diverges!

jhannybean · Answer

the p-series also proves that the harmonic series diverges. the power of n is = 1, $\ p \le 1$ and therefore divergent toooo.

anonymous · Answer

Oh, you're right, I overlooked that. Then the answer is that it converges only for x = 0 and that the sum is 0, right?

anonymous · Answer

Well for $x=0$ then it converges to $0$ yes since it's just $0+0+0+\dots$. For any $x\ne0$ there's no way! Be careful that you're not thinking of $x^{2n}$ or $x^2$ rather than $x^2$ -- P.S. I only checked Wolfram after I answered myself.

jhannybean · Answer

What kind of series is this btw...

anonymous · Answer

I don't doubt that, it's only that I always try everything I know before I post, thanks :)

anonymous · Answer

@Jhannybean just a weird "series"... it's not a power series

jhannybean · Answer

Ohh ok. I thought this was a power series for a minute.

anonymous · Answer

I'd say it's just functional series.

anonymous · Answer

I mean technically you could consider it one if the coefficient converged to a constant value where the coefficients of all powers other than $x^2$ are $0$.