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Is there any formula that directly gives the result of this summation: $$ \sum_{k=-\infty}^n a^k$$ ?

YoTengoUnLCD
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Uygar Uçar
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4 Answers4

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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

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

Olivier Oloa
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    If one divides out $a^{n}$ and rewrites the sum as $a^n \sum_{k=0}^\infty a^{-k}$ then the geometric series is directly applicable (no splitting needed). – Semiclassical Jun 05 '16 at 18:50
  • @Semiclassical Good catch! – Olivier Oloa Jun 05 '16 at 18:51
  • @Semiclassical Another short way: $\sum_{k=-n}^\infty a^{-k}$ (no need to factor out $a^n$) – Olivier Oloa Jun 05 '16 at 19:03
  • My preferred way (which is quite similar to Yves's answer): $$(a-1)\sum_{k=-\infty}^n a^k=\sum_{k=-\infty}^{n+1} a^k-\sum_{k=-\infty}^{n} a^k=a^{n+1}\implies \sum_{k=-\infty}^n a^k=\frac{a^{n+1}}{a-1}$$ – Semiclassical Jun 05 '16 at 19:06
  • Yes, but you are proving the formula... Not just applying. But since it is a short proof, I understand your point of view. – Olivier Oloa Jun 05 '16 at 19:09
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$$ S_{n+1}=\sum_{k=-\infty}^{n+1} a^k=S_n+a^{n+1}$$ and

$$S_{n+1}=\sum_{k=-\infty}^n a^{k+1}=aS_n$$ give

$$S_n=\frac{a^{n+1}}{a-1}.$$

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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.

YoTengoUnLCD
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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 $n$.

mathreadler
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    Erm. It may only be me, but regardless of whether it being correct, this answer makes me orders of magnitude more confused about the question than I was before reading it. – Clement C. Jun 05 '16 at 19:04
  • That depends solely on the reader. I can't know the background of everyone who reads questions and answers. One answer can confuse one reader but may help another. – mathreadler Jun 05 '16 at 19:07
  • Well, admittedly. But from your answer: where does $n$ come up in what you write, for instance? (To ask for a specific point.) (Also, sorry for the grammatical mistake I made above.) – Clement C. Jun 05 '16 at 19:08
  • Tried making it a bit more clear with an example in matlab. $n$ is the position of the last $1$ in the integral filter. – mathreadler Jun 05 '16 at 19:22