Grandi's series and imaginary units

Question

First of all, I am not a mathematician, so be a little indulgent.

Let's begin from $S(x) = \sum_{n=0}^{\infty}x^n $

We know that under the usual condition $|x|<1$ this summation is a geometric series that converges to the fraction $\frac{1}{1-x}$. As far I know this rule holds for complex numbers also. A similar formula works for squared matrix, if $|A|<1$ then $I + A + {A}^{2} + ...$ converges to ${(I-A)}^{-1}$ (for example: Understanding the Leontief inverse).

Now let's write Grandi's infinite series ($1 - 1 + 1 - 1 + ...$) as $G := \sum_{n=0}^{\infty}(-1)^n$ and let's take the sum $H = G + Gi$, where $i$ is the imaginary unit. So, $H$ is just a modified Grandi's series and we may begin write it down like $1 + i - 1 - i + 1 + i - 1 - i + ...$ and so on.

Then, in this form, could $H$ be correctly described as $H = \sum_{n=0}^{\infty}i^n $ ?

If so, could we use the geometric series rule and say that, in the limit, using the $S(x)$ form, as $x$ approaches $i$, $H$ tend to converge to $\frac{1}{1-i}=\frac{1}{2}+\frac{1}{2}i$ ?

If all the above is correct, could we use it to show that $G$ converges to $\frac{1}{2}$ ?

Finally, is this mixing up imaginary units and infinite sums a sound reasoning ?

Thanks.

Matteo.

Answer 1

Geometric series $\sum_n z^n$ are perfectly fine for complex numbers. They converge if and only if $|z| < 1$. In your examples $G$ and $H$, $|z| = 1$ so they do not converge.

This is not, however, the end of the matter. You might look up e.g. Abel summation. The Abel summations of $G$ and $H$ are indeed $\frac{1}{2}$ and $\frac{1}{2} + \frac{i}{2}$ respectively.

Answer 2

You can use complex analysis to deal with such convergence problems. The series has a radius of convergence of 1, because the function the series converges to, $\frac{1}{1+z}$ has a singularity at $z= -1$ which is at a distance of $1$ from the origin. The radius of convergence is equal to this distance. So, even if you move from the origin in the positive $x$-direction, opposite to where the singularity is, your series is going to start to diverge as soon as you pass the point $z = 1$.

There is a remedy to this problem which works by using a mapping that moves the singularity at $z = -1$ farther from the origin. If we put $z = \frac{w}{2-w}$, then $z = 1$ corresponds to $w = 1$, however, in the $w$-plane the point $w = 1$ now lies within the radius of convergence. In fact, as a function of $w$ the singularity has completely vanished as the function has become $1-\frac{w}{2}$. For $w = 1$ this is $\frac{1}{2}$ which is consistent with the Abel summation result.

This particular transform is known as the Euler transform. In general, replacing $z$ by $g(w)$ such that $g(0) = 0$ and $g(1) = 1$ will allow you re-expand the series without knowing what function it sums to. You then want to choose this function such that the nearest singularity to the origin lies outside the circle $|w|=1$.