In a problem I have both $\sum_{n=0}^{\infty}{a^n}$ and $\sum_{n=0}^{S}{a^n}$ (finite and infinite). To simplify I found that the infinite summation is:
$$\sum_{n=0}^{\infty}{a^n} = \dfrac{1}{1-a}$$
But for the finite summation I found two different results (notice that both start at zero):
Here: $\sum_{n=0}^{S}{a^n} = \dfrac{a^{S+1}-1}{a-1} \tag{1}$
And here: $\sum_{n=0}^{S}{a^n} = \dfrac{1-a^{S+1}}{1-a} \tag{2}$ (mathworld.wolfram.com/GeometricSeries.html)
($a$ fulfills the condition $|a|\lt 1$)
According to my math the two are not equivalent, Which is the correct one?
I also need formulas for both, finite and infinite, of this serie $\sum_{n=0}{n \cdot a^n}$ (both starting at zero). In wolframalpha I found that:
$$\sum_{n=0}^{\infty}{n \cdot a^n} = \dfrac{a}{(a-1)^2} \tag{3}$$
and
$$\sum_{n=0}^{S}{n \cdot a^n} = \dfrac{(a \cdot S - S - 1) \cdot a^{S + 1} + a}{(1-a)^2} \tag{4}$$
Are these correct? And, how are these last series called? I'd like to see a proof but I didn't know the name to look for.
Thanks.
EDIT: I fixed (1).
$\sum_{n=0}^{N}{a^n} = \dfrac{1-a^{N+1}}{1-a}$ This is true as long as $a \ne 1$.
Now consider $|a|<1$, and let $N \rightarrow \infty$.
Then $a^N \rightarrow 0$, so
$\lim_{N \rightarrow \infty} \sum_{n=0}^{N}{a^n} = \dfrac{1-a^{N+1}}{1-a}=\dfrac{1}{1-a}.$
The last variations can be shown by induction or by using calculus.
[1] is wrong, but you can fix it by starting the summation at $n=1$ instead of $n=0$.
$\sum_{n=1}^{S}{a^n} = \dfrac{a (a^S-1)}{(a-1)}$.