Power Series Sum: Summing (x^n/(n²+2n)) from n=1 to ∞

[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.

Anyone an idea?

 

[dextercioby]

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.

Anyone an idea?

1.Factor the denominator:

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

Write your series like:
$$S(x)=\frac{1}{2}(\sum_{n=1}^{+\infty} \frac{x^{n}}{n}-\sum_{n=1}^{+\infty} \frac{x^{n}}{n+2})$$

Daniel.

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

 

[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..

 

[NateTG]

1.Factor the denominator:

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

Write your series like:
$$S(x)=\frac{1}{2}(\sum_{n=1}^{+\infty} \frac{x^{n}}{n}-\sum_{n=1}^{+\infty} \frac{x^{n}}{n+2})$$

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.

 

[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.

 

[NateTG]

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

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.)