find a closed-form expression for sum(n*(n+1)/2^n)

[johnk]

Hello,

I need help finding a closed-form expression for the following sum:
\(\displaystyle \sum_{n=1}^{\infty}\frac{{n\cdot (n+1)}}{2^n}\)

I tried...
\(\displaystyle \lim_{x \to 1-} \sum_{n=1}^{\infty}\frac{{n\cdot (n+1)}}{2^n}\cdot x^{2^n} = \lim_{x \to 1-} \int \sum_{n=1}^{\infty} n\,\left( n+1\right) \,{x}^{{2}^{n}-1} dx\)
... to get rid of the fraction, but it doesn't seem to get me any closer to anything I can sum.
All exercises I've done before were relatively easy to manipulate into something like \(\displaystyle \small \sum x^n\), but I can't figure this one out.

 

[pka]

You know that \(\displaystyle \frac{1}{{1 - x}} = \sum\limits_{k = 0}^\infty {x^k } ,\quad \left|x\right| < 1\).
From which it follows that \(\displaystyle y=\frac{x}{{1 - x}} = \sum\limits_{k = 0}^\infty {x^{k+1} } ,\quad \left|x\right| < 1\).
\(\displaystyle {xy''}=\frac{2x}{{(1 - x)}^3} = \sum\limits_{k = 0}^\infty {k(k+1)x^{k} } ,\quad \left|x\right| < 1\).

Now can you finish, \(\displaystyle x=\frac{1}{2}\)?

 

[johnk]

Thanks, pka.

\(\displaystyle {xy''}=\frac{2x}{{(1 - x)}^3} = \sum\limits_{k = 0}^\infty {k(k+1)x^{k} } ,\quad \left|x\right| < 1\)

This looks quite close to where I got. Should I be trying to somehow change it further to \(\displaystyle \sum\limits_{k = 0}^\infty {k(k+1)x^{2^k-1} }\), as I arrived to? Or was I completely off with my first attempt and this is a different approach?

I don't understand at all why x = 1/2

 

[pka]

Is it clear to you that \(\displaystyle \sum\limits_{k = 0}^\infty {k\left( {k + 1} \right)x^k } = \sum\limits_{k = 1}^\infty {k\left( {k + 1} \right)x^k }\)? WHY?

\(\displaystyle \sum\limits_{k = 1}^\infty {k\left( {k + 1} \right)x^k } ,\quad x = \frac{1}{2} \Rightarrow \quad \sum\limits_{k = 1}^\infty {\frac{{k\left( {k + 1} \right)}}{{2^k }}}\)

 

[johnk]

Oh... That was easier than I expected. I acutally did that, but when I arrived at
\(\displaystyle \sum\limits_{k = 0}^\infty k\cdot (k+1)\cdot\frac{1}{2^k}} = \frac{y''}{2} = \frac{1}{(1-\frac{1}{2})^3} = 8\)
I thought that can't be right... Now I see why it is.

Is it clear to you that \(\displaystyle \sum\limits_{k = 0}^\infty {k\left( {k + 1} \right)x^k } = \sum\limits_{k = 1}^\infty {k\left( {k + 1} \right)x^k }\)? WHY?

That's clear, the first term (k = 0) is multiplied by zero.

Thank you for your help, pka.