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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

5 Answers 5

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
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@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

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): $$\begin{align*} \sum_{k=0}^{\infty}\frac{i^k}{k!} &= 1+\frac{i}{1!}+\frac{i^2}{2!}+\frac{i^3}{3!}+\frac{i^4}{4!}+\frac{i^5}{5!}+\cdots \\ &= 1+\frac{i}{1!}+\frac{-1}{2!}+\frac{-i}{3!}+\frac{1}{4!}+\frac{i}{5!}+\cdots \\ &=\left(1+\frac{-1}{2!}+\frac{1}{4!}+\cdots\right)+\left(\frac{i}{1!}+\frac{-i}{3!}+\frac{i}{5!}+\cdots\right) \\ &=\left(1-\frac{1}{2!}+\frac{1}{4!}+\cdots\right)+i\left(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}+\cdots\right) \\ &=\cos(1)+i\cdot\sin(1)\\ \end{align*}$$

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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
    
@Alyosha There are so many different ways of getting at these formulae that I'm not sure the concepts of 'backwards' or 'forwards' even make sense here. That said, I don't know of any proof of that expansion that even involves sin/cos, so I'm not sure how it could be going backwards per se. –  Steven Stadnicki Jan 2 '13 at 21:27

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