Let us revisit the geometric progression sum considered in an earlier article,

s_r = \sum_{k=0}^\infty r^k = 1 + r + r^2 + r^3 + \ldots,

where r here is a complex number. For what values of r does this infinite sum make sense? Can we find a closed-form expression for s_r in such cases? To investigate this, we fix r to some value and consider the partial sums:

s_r(n) = \sum_{k=0}^n r^k = 1 + r + r^2 + \ldots + r^{n-1},

where we just add the first n terms of s_r. Now if s_r(n) tends to a finite limit v as n \rightarrow \infty (can we for any \delta > 0 find an n_0 such that |v-s_r(n)| \leq \delta for all n \geq n_0?) then we have s_r = v.

Let us first single out the special case r=1. Since s_r(n) = n we cannot assign any well-defined, finite value to s_1, so s_r is divergent for r=1. For r \neq 1 we get

(1-r) s_r(n) = 1-r^n \quad \Longleftrightarrow \quad s_r(n) = \frac{1-r^n}{1-r}.

Let us consider three different cases. If |r| < 1 we see that the only term that depends on n tends to zero so we suspect that the limit is 1/(1-r),

\left| \frac{1}{1-r} – \frac{1-r^n}{1-r} \right| = \left| \frac{r^n}{1-r} \right| = \frac{|r|^n}{|1-r|}.

Since the magnitude of the difference between our suspected limit and the partial sums can be made as small as we like (as long as we choose n sufficiently large), we have

s_r = \frac{1}{1-r}, \quad \hbox{for } |r| < 1.

What about |r| > 1? We get

|s_r(n)| = \left| \frac{1-r^n}{1-r} \right| \geq \frac{|r^n|-1}{|r-1|} \geq \frac{|r|^n-1}{|r|+1},

and we see that |s_r(n)| \rightarrow \infty as n \rightarrow \infty. We can thus not find a finite limit to which s_r(n) tends as n \rightarrow \infty, so the series s_r is divergent for |r| > 1.

Left to consider is the case |r|=1, r \neq 1, and this is where it gets interesting. We get

|s_r(n)| = \left| \frac{1-r^n}{1-r} \right| = \frac{|1-r^n|}{|1-r|}
\leq \frac{1 + |r^n|}{|1-r|} = \frac{2}{|1-r|}.

So the partial sums s_r(n) are bounded by some constant independent of n. Does the value 1/(1-r) work as a limit in this case also? We set r = e^{i \theta} with 0 < \theta < 2\pi and subtract,

\begin{aligned}
s_r(n) – \frac{1}{1-r} &= \frac{e^{i \theta n}}{e^{i \theta}-1} = \frac{e^{i \theta n} (e^{-i \theta}+1)}{(e^{i \theta}-1)(e^{-i \theta}+1)} = \frac{e^{i \theta (n-1)} + e^{i \theta n}}{2i\sin \theta} \\
&= \frac{e^{i(\theta(2n-1)-\pi)/2}}{2\sin(\theta/2)}
\end{aligned}

(using e^{i x} + e^{i y} = 2 \cos((x-y)/2) e^{i(x+y)/2} and \sin \theta = 2\sin(\theta/2)\cos(\theta/2)). So s_r(n) does not converge to 1/(1-r) as n \rightarrow \infty. Indeed, we see that s_r(n) follows a circle in the complex plane; a circle centered in 1/(1-r) with radius 1/(2\sin(\theta/2)). And this is what I find interesting: s_r(n) does not converge to any value,

\hbox{the series } s_r \hbox{ is divergent for } |r| \geq 1,

but circles around the value 1/(1-r) when |r|=1, r \neq 1. In fact, 1/(1-r) makes sense for all r \neq 1, so can this value be assigned to s_r in some meaningful way? (When |r| < 1, I would suspect that the values of s_r(n) spirals inward towards 1/(1-r) as n grows and spirals outwards when |r| > 1; I have not verified this, though.)

Hardy: Divergent Series

This reminded me that G. H. Hardy has written a book called Divergent Series, where he manipulates infinite series with an “entirely uncritical spirit”. Therein, he also considers the series s_r and, e.g., s_{-1} = 1/2 can somehow make sense. I have only flicked through the book (excerpt), but I think I should take a closer look…

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