# Convergence of a spiral in $\mathbb{C}$

Does the series $$\sum_{k=0}^{\infty}\frac{i^k}{k!}$$converge, and if so, what is the value of it?

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Do you know the Taylor expansion of $\exp (x)$? –  Fabian Jan 2 '13 at 19:14
Yes, for some reason I forgot it. Thank you all. –  Alyosha Jan 2 '13 at 19:15
This question is so trivial I think I may delete it. –  Alyosha Jan 2 '13 at 19:16
Go ahead....... –  Fabian Jan 2 '13 at 19:17

Hint: $$e^z=\sum_{k=0}^{\infty}\frac{z^k}{k!}$$

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That seems more like an answer than a hint. –  robjohn Jan 2 '13 at 19:56
@robjohn Sometimes the line between answers and hints can be very subtle. This is one of them. –  Nameless Jan 2 '13 at 19:59

The sum of the series is $\exp{(i)}$, which has the value of $\cos{1} + i \sin{1}$ (the arguments being in radians).

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Well, first of all: what's that power series' convergence radius? $$a_k:=\frac{i^k}{k!}\Longrightarrow \frac{a_{k+1}}{a_k}=\frac{i^{k+1}}{(k+1)!}\frac{k!}{i^k}=\frac{i}{k+1}\xrightarrow[k\to\infty]{}0$$

Thus, the series converges for all $\,i\in\Bbb R\,$

Advice: Don't use the letter $\,i\,$ as it usually stands for $\,i=\sqrt{-1}\,$ in mathematics..unless you really meant this $\,i\,$ , of course.

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Recall: $\;\;$for $\large\;x \in \mathbb{C}\,:$ $\large\;\;\;\displaystyle \sum_{k=0}^{\infty}\frac{x^k}{k!} = e^x$

• In your case, we have $\large x = i,\;$ giving us $\large \;e^x = e^i = e^{i\theta},\text{ where}\;\;\theta = 1$.

• Now recall Euler's Formula: $$\large \;\;e^{i\theta} = \cos \theta + i \sin\theta,$$
$\quad$ and simply evaluate at $\;\large\theta = 1$.

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and $e^i=\cos(1)+i\sin(1)$. –  pbs Jan 2 '13 at 20:09
Alyosha: is this clear now? Just checking. $\quad$:-) –  amWhy Jan 3 '13 at 18:04
yes, much better. –  pbs Jan 7 '13 at 10:25
$i^0=1, i^1=i, i^2=-1, i^3=-i, i^4=i^0=1$. Therefore, the numerators of the power series are periodic of period 4; what's more, they split off naturally into a real part (the even members) and an imaginary part (the odd members): \begin{align*} \sum_{k=0}^{\infty}\frac{i^k}{k!} &= 1+\frac{i}{1!}+\frac{i^2}{2!}+\frac{i^3}{3!}+\frac{i^4}{4!}+\frac{i^5}{5!}+\cdots \\ &= 1+\frac{i}{1!}+\frac{-1}{2!}+\frac{-i}{3!}+\frac{1}{4!}+\frac{i}{5!}+\cdots \\ &=\left(1+\frac{-1}{2!}+\frac{1}{4!}+\cdots\right)+\left(\frac{i}{1!}+\frac{-i}{3!}+\frac{i}{5!}+\cdots\right) \\ &=\left(1-\frac{1}{2!}+\frac{1}{4!}+\cdots\right)+i\left(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}+\cdots\right) \\ &=\cos(1)+i\cdot\sin(1)\\ \end{align*}
OK, but isn't that just 1 step backwards in the proof of $e^x$'s MaClaurin expansion? –  Alyosha Jan 2 '13 at 21:12