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?
 A: Hint:
$$e^z=\sum_{k=0}^{\infty}\frac{z^k}{k!}$$
A: 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.
A: The sum of the series is $\exp{(i)}$, which has the value of $\cos{1} + i \sin{1}$ (the arguments being in radians).
A: 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 $\;\large\theta = 1$.
A: Alternately (and equivalently to several of the other answers), if you didn't know the Euler formula but did know the power series for sin and cos, you could reason as follows:
$i^0=1, i^1=i, i^2=-1, i^3=-i, i^4=i^0=1$.  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):
$$\begin{align*}
\sum_{k=0}^{\infty}\frac{i^k}{k!} &= 1+\frac{i}{1!}+\frac{i^2}{2!}+\frac{i^3}{3!}+\frac{i^4}{4!}+\frac{i^5}{5!}+\cdots \\
&= 1+\frac{i}{1!}+\frac{-1}{2!}+\frac{-i}{3!}+\frac{1}{4!}+\frac{i}{5!}+\cdots \\
&=\left(1+\frac{-1}{2!}+\frac{1}{4!}+\cdots\right)+\left(\frac{i}{1!}+\frac{-i}{3!}+\frac{i}{5!}+\cdots\right) \\
&=\left(1-\frac{1}{2!}+\frac{1}{4!}+\cdots\right)+i\left(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}+\cdots\right) \\
&=\cos(1)+i\cdot\sin(1)\\
\end{align*}$$
