A "spiral" in the Complex plane

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Homework Statement
Starting from the origin, go one unit east, then the same length north, then (1/2) of the previous length west, then (1/3) of the previous length south, then (1/4) of the previous length east, and so on. What point does this “spiral” converge to?
Relevant Equations
series sum
I understand that the "spiral" converges to 1+i-1/2-i/3!+1/4!+i/5!-1/6!-i/7!... .
It splits into two: one for Re, 1-1/2+1/4!-1/6!..., and the other for Im, 1-1/3!+1/5!-1/7!... .
Any hints on how to compute them?
 
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What are <br /> \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!} and \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} when x = 1?
 
pasmith said:
What are <br /> \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!} and \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} when x = 1?
##\cos## and ##\sin##, of course. Thanks!
 
Or, even better:
$$
\sum_{n=0}^\infty \frac{(ix)^n}{n!}
$$
 
Orodruin said:
Or, even better:
$$
\sum_{n=0}^\infty \frac{(ix)^n}{n!}
$$
Yes. Straight.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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