Summation $ \sum_{k=-\infty}^n a^k$ Is there any formula that directly gives the result of
this summation:
$$
\sum_{k=-\infty}^n a^k$$
?
 A: One may write
$$
\sum_{-\infty}^na^k=\sum_{-\infty}^0 a^k+\sum_1^n a^k
$$ then one may use the standard geometric series
$$
\sum_{k=0}^n x^k=\frac{1-x^{n+1}}{1-x}, \,\, x\neq1,\quad \sum_{k=0}^\infty x^k=\frac{1}{1-x},\quad |x|<1.
$$
Finally the sought sum is

$$
\sum_{-\infty}^na^k=\sum_0^{\infty} a^{-k}+\sum_1^n a^k=\frac{1}{1-\frac1a}+\frac{a(1-a^{n})}{1-a}=\frac{a^{n+1}}{a-1},\quad |a|>1.
$$

A: $$
S_{n+1}=\sum_{k=-\infty}^{n+1} a^k=S_n+a^{n+1}$$
and
$$S_{n+1}=\sum_{k=-\infty}^n a^{k+1}=aS_n$$ give
$$S_n=\frac{a^{n+1}}{a-1}.$$
A: Hint: separate the sum, from $-\infty$ to $0$, and from $0$ to $n$. Make the change of variables $k\mapsto -k$ in the first one.
A: Another aspect to view this from is convolution/deconvolution pair of a discretized differential operator $[-1,1]$ has $\mathcal{Z}$ transform $a-1$. And the integral filter which is $[\cdots,1,0,\cdots]$ would have the given series as it's $\mathcal{Z}$-transform (here $n$ is the position of the last 1). The integral of the derivative is the identity [1] and as composition corresponds to multiplication in the $\mathcal{Z}$ domain we have our correspondence between the three of them.
We can check by a discrete approximation of the convolution in matlab or octave:
a = [1,1,1,1,0,0,0,0]; b = [0,-1,1];
conv(a',b','valid')'
ans =
0   0   0   1   0   0

a is the integral of the first four values, b is a differential approximation. In theory a would be of infinite length of 1s from $-\infty$ up to position $n$.
