janmr blog

On the Divergence of a Geometric Progression Sum

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

sr=k=0rk=1+r+r2+r3+,s_r = \sum_{k=0}^\infty r^k = 1 + r + r^2 + r^3 + \ldots,

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

sr(n)=k=0nrk=1+r+r2++rn1,s_r(n) = \sum_{k=0}^n r^k = 1 + r + r^2 + \ldots + r^{n-1},

where we just add the first nn terms of srs_r. Now if sr(n)s_r(n) tends to a finite limit vv as nn \rightarrow \infty (can we for any δ>0\delta > 0 find an n0n_0 such that vsr(n)δ|v-s_r(n)| \leq \delta for all nn0n \geq n_0?) then we have sr=vs_r = v.

Let us first single out the special case r=1r=1. Since sr(n)=ns_r(n) = n we cannot assign any well-defined, finite value to s1s_1, so srs_r is divergent for r=1r=1. For r1r \neq 1 we get

(1r)sr(n)=1rnsr(n)=1rn1r.(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|r| < 1 we see that the only term that depends on nn tends to zero so we suspect that the limit is 1/(1r)1/(1-r),

11r1rn1r=rn1r=rn1r.\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 nn sufficiently large), we have

sr=11r,for r<1.s_r = \frac{1}{1-r}, \quad \text{for } |r| < 1.

What about r>1|r| > 1? We get

sr(n)=1rn1rrn1r1rn1r+1,|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 sr(n)|s_r(n)| \rightarrow \infty as nn \rightarrow \infty. We can thus not find a finite limit to which sr(n)s_r(n) tends as nn \rightarrow \infty, so the series srs_r is divergent for r>1|r| > 1.

Left to consider is the case r=1|r|=1, r1r \neq 1, and this is where it gets interesting. We get

sr(n)=1rn1r=1rn1r1+rn1r=21r.|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 sr(n)s_r(n) are bounded by some constant independent of nn. Does the value 1/(1r)1/(1-r) work as a limit in this case also? We set r=eiθr = e^{i \theta} with 0<θ<2π0 < \theta < 2\pi and subtract,

sr(n)11r=eiθneiθ1=eiθn(eiθ+1)(eiθ1)(eiθ+1)=eiθ(n1)+eiθn2isinθ=ei(θ(2n1)π)/22sin(θ/2)\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 eix+eiy=2cos((xy)/2)ei(x+y)/2e^{i x} + e^{i y} = 2 \cos((x-y)/2) e^{i(x+y)/2} and sinθ=2sin(θ/2)cos(θ/2)\sin \theta = 2\sin(\theta/2)\cos(\theta/2)). So sr(n)s_r(n) does not converge to 1/(1r)1/(1-r) as nn \rightarrow \infty. Indeed, we see that sr(n)s_r(n) follows a circle in the complex plane; a circle centered in 1/(1r)1/(1-r) with radius 1/(2sin(θ/2))1/(2\sin(\theta/2)). And this is what I find interesting: sr(n)s_r(n) does not converge to any value,

the series sr is divergent for r1,\text{the series } s_r \text{ is divergent for } |r| \geq 1,

but circles around the value 1/(1r)1/(1-r) when r=1|r|=1, r1r \neq 1. In fact, 1/(1r)1/(1-r) makes sense for all r1r \neq 1, so can this value be assigned to srs_r in some meaningful way? (When r<1|r| < 1, I would suspect that the values of sr(n)s_r(n) spirals inward towards 1/(1r)1/(1-r) as nn grows and spirals outwards when r>1|r| > 1; I have not verified this, though.)

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 srs_r and, e.g., s1=1/2s_{-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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