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I have a summation $\sum\limits_{k = -\infty}^{\infty} r^k$ that I've split up to $\sum\limits_{k = -\infty}^{-1} r^k + \sum\limits_{k = 0}^{\infty} r^k$. The second summation is a solvable geometric series, which gives me $\frac{1}{1 - r}$.

To compute the second summation, I am trying to change the variable index and use the same geometric series. I set $a = -k$ and use $\sum\limits_{a = 0}^{\infty} r^a$. This over-counts by one, but subtracting out the zero term at the end is easy.

My question is, how does this change of variable affect the value with respect to $k$? Surely this half of the summation is not also equal to $\frac{1}{1 - r}$ (- 1 to account for the extra 0 term). What would its value be instead?

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The problem is, it becomes $\sum_{a=1}^{\infty} (\frac{1}{r})^{a}$, and if $0 \leq r < 1$ (which is necesary for the "other half" to converge for real non-negative $r$), then $\frac{1}{r} > 1,$ and this half now diverges. – Geoff Robinson Mar 22 '12 at 21:18
(+1). And don't call me Shirley. – The Chaz 2.0 Mar 22 '12 at 21:33
Does this question come from another one, about maybe trigonometric functions, or hyperbolic functions? If so, you could post the original question, since the two ways infinite geometric progression approach doesn't work. – André Nicolas Mar 22 '12 at 21:50
You can use $\frac{1}{1-r} = 1 + r + r^2 + \dots$ and $\frac{1}{1-(1/r)} = 1 + r^{-1} + r^{-2} + \dots$ together by summing these and subtracting 1. I believe the result is 0, which is (perhaps paradoxically) the sum of this divergent series. There's a webpage devoted to this topic at . – Matt Groff Mar 22 '12 at 22:05
up vote 1 down vote accepted

You can see which change is required by considering a partial sum: $$\sum_{k = -n}^{-1} r^k = r^{-n} + r^{-(n-1)} + \cdots + r^{-1} = \sum_{j=1}^n r^{-j}$$ Therefore, you can calculate this sum by noting that $r^{-j} = (r^{-1})^j$ (i.e., the geometric series sums to $\frac{r^{-1}}{1-r^{-1}}$ under the assumption that the series converges. But note that the geometric series $\sum_{k=0}^\infty a^k$ converges only if $|a| < 1$. Hence, only one half of your series converges, while the other diverges. In particular, $\sum_{k=-\infty}^\infty r^k$ will always diverge, namely against $\infty$ if $r > 0$ and indefinitely otherwise.

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