Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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: $$e^z=\sum_{k=0}^{\infty}\frac{z^k}{k!}$$

share|cite|improve this answer
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).

share|cite|improve this answer

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.

share|cite|improve this answer

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 $\;\large\theta = 1$.

share|cite|improve this answer
and $e^i=\cos(1)+i\sin(1)$. – poirot Jan 2 '13 at 20:09
Alyosha: is this clear now? Just checking. $\quad$:-) – amWhy Jan 3 '13 at 18:04
yes, much better. – poirot 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): $$\begin{align*} \sum_{k=0}^{\infty}\frac{i^k}{k!} &= 1+\frac{i}{1!}+\frac{i^2}{2!}+\frac{i^3}{3!}+\frac{i^4}{4!}+\frac{i^5}{5!}+\cdots \\ &= 1+\frac{i}{1!}+\frac{-1}{2!}+\frac{-i}{3!}+\frac{1}{4!}+\frac{i}{5!}+\cdots \\ &=\left(1+\frac{-1}{2!}+\frac{1}{4!}+\cdots\right)+\left(\frac{i}{1!}+\frac{-i}{3!}+\frac{i}{5!}+\cdots\right) \\ &=\left(1-\frac{1}{2!}+\frac{1}{4!}+\cdots\right)+i\left(\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}+\cdots\right) \\ &=\cos(1)+i\cdot\sin(1)\\ \end{align*}$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.