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## Main Question or Discussion Point

Consider the infinite sum:

[itex]\sum_{n=1}^\infty \frac{n^2}{2^n}[/itex]

For the impatient of you, the answer is here.

Anyways, I'm trying to generalize this result, so let me state a definition:

[itex]\sigma_\alpha(k) = \sum_{n=1}^\infty n^k \alpha ^ n[/itex]

This sum converges so long as the magnitude of [itex]\alpha[/itex] is less than 1. I won't prove that; moreover (and I won't bother with the derivation because it involves typing too much LaTeX on my part) I found a way to get solutions to this:

[itex]\sigma_\alpha(k) = \left(\frac{\alpha}{1-\alpha}\right)\left[1 + \sum_{l=0}^{k-1} \binom {k}{l} \sigma_\alpha(l)\right][/itex]

Let's take the cleanest case, where [itex]\alpha={1/2}[/itex].

[itex]\sigma_{1/2}(k) = 1 + \sum_{l=0}^{k-1} \binom {k}{l} \sigma_{1/2}(l)[/itex]

Then [itex]\sigma_{1/2}(2)[/itex] should give us the answer above, and it does.

So my question:

It would be much easier if there wasn't the "1 + "...

Cheers,

Andrew.

[itex]\sum_{n=1}^\infty \frac{n^2}{2^n}[/itex]

For the impatient of you, the answer is here.

Anyways, I'm trying to generalize this result, so let me state a definition:

[itex]\sigma_\alpha(k) = \sum_{n=1}^\infty n^k \alpha ^ n[/itex]

This sum converges so long as the magnitude of [itex]\alpha[/itex] is less than 1. I won't prove that; moreover (and I won't bother with the derivation because it involves typing too much LaTeX on my part) I found a way to get solutions to this:

[itex]\sigma_\alpha(k) = \left(\frac{\alpha}{1-\alpha}\right)\left[1 + \sum_{l=0}^{k-1} \binom {k}{l} \sigma_\alpha(l)\right][/itex]

Let's take the cleanest case, where [itex]\alpha={1/2}[/itex].

[itex]\sigma_{1/2}(k) = 1 + \sum_{l=0}^{k-1} \binom {k}{l} \sigma_{1/2}(l)[/itex]

Then [itex]\sigma_{1/2}(2)[/itex] should give us the answer above, and it does.

So my question:

**is there a way to go from my recursive expression for [itex]\sigma_{1/2}(k)[/itex] to a non-recursive formula, just in terms of k?**It would be much easier if there wasn't the "1 + "...

Cheers,

Andrew.