2
$\begingroup$

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?

  • 0
    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 http://cornellmath.wordpress.com/2007/07/28/sum-divergent-series-i/ .2012-03-22

1 Answers 1

2

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.