Formula for weighed geometric sum

Question

I'm trying to find an easy way to derive a formula for:

$S_{n} = \frac{1}{n}\sum_{i=0}^{n}(n-i)x^{i}$

I've found a recurrence relationship of sorts:

$S_{n+1} = \frac{xnS_{n}+n+1}{n+1} = x\frac{n}{n+1}S_{n}+1$

But also:

$S_{n+1} = \frac{1}{n+1}\sum_{i=0}^{n+1}(n+1-i)x^{i}$

$S_{n+1} = \frac{1}{n+1}\sum_{i=0}^{n}(n+1-i)x^{i}$ (as where $i=n+1$, $n+1-i=0$)

$S_{n+1} = \frac{1}{n+1}\sum_{i=0}^{n}(n-i)x^{i} + \frac{1}{n+1}\sum_{i=0}^{n}x^{i}$

$S_{n+1} = \frac{n}{n+1}S_{n} + \frac{1}{n+1}\frac{1-x^{n+1}}{1-x}$

So equating these two:

$x\frac{n}{n+1}S_{n}+1 = \frac{n}{n+1}S_{n} + \frac{1}{n+1}\frac{1-x^{n+1}}{1-x}$

$xnS_{n} + (n+1) = nS_{n}+\frac{1-x^{n+1}}{1-x}$

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

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

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

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

$S_{n} = \frac{n-(1-x^{n})}{n{(1-x)}}$

$S_{n} = \frac{1-\frac{1}{n}(1-x^{n})}{1-x}$

Or

$S_{n} = \frac{1}{1-x}-\frac{1-x^{n}}{n(1-x)}$

So if $|x|<1$, $S_{n} \rightarrow \frac{1}{1-x}$ as $n \rightarrow \infty$

Questions:

• I appear to be making heavy weather of this (certainly the finite sum); is there an easier way?

• Is this series well known?

Answer 1

Lets set $G_n = \sum_{i = 0}^{n}x^i$, then: $$ S_n = G_n - \frac{x}{n}\frac{\partial}{\partial x}G_n$$ From here we can derive finite and infinite sums - assuming, of course, convergence and possibility to swap sum and differentiation.