Why $e^{-n\pi i} + (-1)^{n}i$ is divergent?

Question

This is not a homework question.

The series in question is: $$z_{n} = e^{-n\pi i} + (-1)^n i$$

Simplification would results into: $$ z_{n} = \cos(n\pi)+ (-1)^n i$$

and I think neither of these limits, $lim_{n \to \infty} \cos(n\pi)$ and $ \lim_{n \to \infty} (-1)^n$, exists. Correct?

If they are taken as limits in $n$, are there subsequences which have different limit values? — abiessu, Aug 21 '15 at 12:33
The sum of two divergent sequences can still converge. Consider $a_n=(-1)^n$ and $b_n=\frac1n-(-1)^n.$ — Cameron Buie, Aug 21 '15 at 12:41
@abiessu Yes. for $cos(n\pi)$, we have $a_n = {1}$ and $b_n = {-1}$ and same would act as a sub-sequences for $(-1)^n$ — Sohail, Aug 21 '15 at 12:42
@HansLundmark, I was asked of existence of sub-sequences and I answered. — Sohail, Aug 21 '15 at 12:50
I was referring to the formulation of the original question (and the tag divergent-series, which seems out of place). — Hans Lundmark, Aug 21 '15 at 13:18

score 2 · Accepted Answer · answered Aug 21 '15 at 12:43

2

Hint:

If $a_n,b_n\in\mathbb R$ then $(a_n+ib_n)_n$ will converge if and only if $(a_n)_n$ and $(b_n)_n$ both converge.

answered Aug 21 '15 at 12:43

drhab

151,093

I could not figure out the use of this result, as in a present case I have two divergent series and sum of two divergent series can still converge. Where does this results of your come in? – Sohail Aug 21 '15 at 13:02
1

Yes, if $u_n,v_n\in\mathbb C$ and $u_n$ and $v_n$ both diverge, then still it is possible that $u_n+v_n$ converges (e.g. if $v_n=-u_n$). But not if the $u_n$ are real (like $a_n$ in my answer) and the $v_n$ are imaginary (like the $ib_n$ in my answer) numbers. The fact that $\cos(n\pi)$ is not converging is allready enough to conclude that there is no convergence. – drhab Aug 21 '15 at 17:12

score 0 · Answer 2 · answered Aug 21 '15 at 14:12

Starting from the beginning where we take $n\in\Bbb Z$ with

$$z_{n} = e^{-n\pi i} + (-1)^n i\tag 1$$

we have $e^{-n\pi i}=\cos(n\pi)-i\sin(n\pi)=\cos(n\pi)-i\cdot 0$, so we can rewrite $(1)$ as

$$z_n=\cos(n\pi)+(-1)^ni\tag 2$$

Now we have two sequences, $a_n=\cos(n\pi)=1,-1,1,-1,\dots$ and $b_n=(-1)^ni=i,-i,i,-i,\dots$ Individually, neither sequence converges. Summed together, the result (i.e., $z_n$) cannot converge, since we have

$$z_n=\cos(n\pi)+(-1)^ni=1+i,-1-i,1+i,-1-i,\dots$$

Why $e^{-n\pi i} + (-1)^{n}i$ is divergent?

2 Answers2