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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.

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    $\begingroup$ Technically... you should have $$H=(1-1+1-1+\dots)+(i-i+i-i+\dots)$$and one would not usually combine the two until they knew what these meant as far as Cesàro sum or Abel sum or whatever, then you can decide if combining these two would be reasonably doable. $\endgroup$ Commented Mar 30, 2017 at 21:27

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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 $|w|=1$.

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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.

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  • $\begingroup$ I agree H does not converge. This the flaw. In my post I didn't mention what I had in mind, that is in the limit, as x approaches $i$ then it tends to reach that value $0.5+0.5i$. Is that a way ? $\endgroup$ Commented Mar 30, 2017 at 21:18
  • $\begingroup$ Yes, it seems I forgot to look up Abel summation. It quite perfectly describes what I was saying. $\endgroup$ Commented Mar 30, 2017 at 21:28

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