# Power serie sum

1. Jan 31, 2005

### bulbanos

sum(x^n/(n²+2n),n=1..infinity)

we've never heard of hypergeom functions and we encountered this on our exam. The question was not only the radius of convergence but the infinite sum as well. :surprised

Anyone an idea?

2. Jan 31, 2005

### dextercioby

1.Factor the denominator:

$$\frac{1}{n^{2}+2n}=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2}})$$ (1)

$$S(x)=\frac{1}{2}(\sum_{n=1}^{+\infty} \frac{x^{n}}{n}-\sum_{n=1}^{+\infty} \frac{x^{n}}{n+2})$$

Daniel.

P.S.I don't know how u can connect your series with the hypergeomtric ones.I'll think about it...

Last edited: Jan 31, 2005
3. Jan 31, 2005

### arildno

Continue by writing:
$$S(x)=\frac{1}{2}(\int_{0}^{x}\sum_{n=1}^{\infty}t^{n-1})dt-\frac{1}{x^{2}}\int_{0}^{x}(\sum_{n=1}^{\infty}t^{n+1})dt))=$$
$$\frac{1}{2}(\int_{0}^{x}\frac{dt}{1-t}-\frac{1}{x^{2}}\int_{0}^{x}\frac{t^{2}dt}{1-t})$$
or something like that..

Last edited: Jan 31, 2005
4. Jan 31, 2005

### NateTG

You're moving terms in a conditionally convergent sequence which is not always kosher. Specifically, in this case, if $x=1$ the original series is convergent, while yours is not.

I can suggest an alternative approach:
As you pointed out:
$$\frac{1}{n^{2}+2n}=\frac{1}{2}(\frac{x^n}{n}-\frac{x^n}{n+2}})$$
This looks suspicously like a telescoping sum. Let's take a look at partial sums
$$S(x,k)=\frac{1}{2}\sum_{n=1}^{k}\left(\frac{x^n}{n}-\frac{x^n}{n+2}}\right)$$
In the $x=1$ this is very nice:
$$S(1,k)=\frac{1}{2}\left(\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...\left(\frac{1}{k}-\frac{1}{k+2}\right)\right)$$
Now, the negative
$\frac{1}{3}$
from the first term will cancel with the positve
$\frac{1}{3}$
from the 3rd term. Similarly, for the negative elements in each of the following terms except for the last two. This means that the sum telescopes to:
$$S(1,k)=\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2}\right)$$
so
$$S(1)=\frac{3}{4}$$

It's pretty obvious that for $|x|<1$ the series are absolutely convergent (so Dex's approach will work) and for $$|x|>1$$ the series will be divergent since the individual terms will grow without bound.

Last edited: Jan 31, 2005
5. Jan 31, 2005

### dextercioby

There's a typo
$$S(1)=\frac{3}{4}$$

What do you mean by
$$|x|<0$$

Isn't the modulus ALWAYS REAL AND NONEGATIVE??

Daniel.

Last edited: Jan 31, 2005
6. Jan 31, 2005

### NateTG

Right, forgot the $\frac{1}{2}$.
[/QUOTE]
What do you mean by
$$|x|<0$$
Isn't the modulus ALWAYS REAL AND NONEGATIVE??[/QUOTE]

I meant to say $|x|<1$ and $|x|>1$.

(Will edit the post.)