Showing that $i^n=0$ when $i$ is the imaginary unit and $n$ is infinite

Question

Let be the sum: $$\begin{align} s&= i+i^2+i^3+\cdots+i^n\\ &= i-1-i+1+i-\cdots\\ &=(1+i)(1-1+1-1+\cdots) \end{align}$$ As Grandi's sum is equal to $1/2$, so: $s=\frac12(1+i)$.

But, $s$ is also a geometric series, so: $s=\displaystyle\frac{(1-i^{(n+1)})}{(1-i)}$

So: $$\frac12(1+i) = \frac{(1-i^{(n+1)})}{(1-i)}$$

Which means: $$1-i^{(n+1)}=\frac{(1+i)(1-i)} 2=1$$

So $i^n=0$ when $n$ is infinite.

$1-1+1-1+ \ldots$ is a divergent series, and not equal to $1/2$. Have a look at https://math.stackexchange.com/questions/2059629/does-grandis-series-diverge-or-is-it-equal-to-frac12 and the linked threads. — Martin R, Oct 26 '20 at 10:43
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — Martin R, Oct 26 '20 at 10:48
You can't mix Cesàro summation and methods for convergent series. See my answer. — Raffaele, Oct 26 '20 at 12:53

score 1 · Accepted Answer · answered Oct 26 '20 at 12:52

The given series is divergent.

If you want to apply Cesàro summation you must proceed as follows $$s_n=\sum _{k=1}^n i^k=\left(\frac{1}{2}-\frac{i}{2}\right) \left(-1+i^n\right)$$ $$a_m=\sum _{n=1}^m s_n=-\frac{1}{2} i \left(i^m-(1+i) m-1\right)$$ and finally $$S=\underset{m\to \infty }{\text{lim}}\frac{a_m}{m}=-\frac{1}{2}+\frac{i}{2}$$

Showing that $i^n=0$ when $i$ is the imaginary unit and $n$ is infinite

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