# Can a complex quantity have both a finite and an infinite part?

Of course, infinity is not a number but sometimes it is convenient to use it in place of a number, and this is very common in calculus, for example when expressing a limit. So I was wondering if it ever makes sense to define a quantity (not sure if this is the most appropriate term) consisting of the sum of a finite real part and an infinite imaginary part or vice-versa.

The specific example I had in mind was the sum of the $2^k$-th roots of $-1$ where $k$ covers the positive integers, i.e.

$$\sqrt[2]{-1}+\sqrt[4]{-1}+\sqrt[8]{-1}+\dots+\sqrt[2^n]{-1}= \sum_{k=1}^n(-1)^{1/2^k}= \sum_{k=1}^n\cos\frac\pi{2^k}+i\sum_{k=1}^n\sin\frac\pi{2^k}$$

The real part obviously diverges as $n$ grows, but it's easy to see that the imaginary part converges, because of $\lim\limits_{k\to\infty}\frac{\sin\left(\pi/2^{k+1}\right)}{\sin\left(\pi/{2^k}\right)}=1/2<1$. In fact, WoframAlpha gives an approximate value of $2.48105$.

So my question is if it would ever make sense to write the limit of the sum as:

$$\lim\limits_{n\to\infty}{\sum_{k=1}^n(-1)^{1/2^k}}\approx\infty+2.48105i$$

Or is this notation just terribly nonsense?

• This may not answer your question, but sometimes it is useful to use notation like $m(x)=\frac{1}{2 \pi i}\int_{c-i\infty}^{c+i\infty}M(x)x^{-s}ds$. In that integral $c\pm i\infty$ is used to specify an integration path for an inverse Mellin transform. Commented Jan 14, 2016 at 1:00

## 1 Answer

I think your notation is sloppy, since $\infty \notin \mathbb{R}$ (and $\infty \notin \mathbb{C}$). However I don't think it's "wrong".