# Finding the sum of a Taylor expansion

I want to find the following sum:

$$\sum\limits_{k=0}^\infty (-1)^k \frac{(\ln{4})^k}{k!}$$

I decided to substitute $x = \ln{4}$:

$$\sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!}$$

The first thing I noticed is that this looks an awful lot like the series expansion of $e^x$:

$$e^x = \sum\limits_{k=0}^\infty \frac{x^k}{k!}$$

The only obstacle is the $(-1)^k$ term. I tried getting rid of it by rewriting:

\begin{align*} \sum\limits_{k=0}^\infty (-1)^k \frac{x^k}{k!} &= 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \frac{x^5}{5!} + \dots\\ &= (1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots) - (x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots)\\ &= \sum\limits_{k=0}^\infty \frac{x^{2k}}{(2k)!} - \sum\limits_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!} \end{align*}

These sums look a lot like the series expansions of $\sin(x)$ and $\cos(x)$:

\begin{align*} \sin(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}\\ \cos(x) &= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!} \end{align*}

However, these sums do have a $(-1)^k$ term, just when I got rid of it! So now I'm stuck. Can someone help me in the right direction?

-

Note that $$e^{-x}=\sum_{k=0}^\infty \frac{(-x)^k}{k!}=\sum_{k=0}^\infty (-1)^k\frac{x^k}{k!}$$

-

You started fine, but then you got sidetracked:

$$\sum\limits_{k\ge 0}(-1)^k \frac{(\ln{4})^k}{k!}=\sum_{k\ge 0}\frac{(-\ln 4)^k}{k!}=e^{-\ln 4}=\frac1{e^{\ln 4}}=\frac14\;.$$

-
21:51:06 vs 21:51:14. Hmmmm... –  Pedro Tamaroff May 24 '13 at 21:52
Please give a more subtle answer for a question with the homework tag the next time: my goal was not to get the right answer (I don't need this site for that), but to understand how to solve such problems. I am still grateful for your help and thus upvoted your answer, but I will accept Peter Tamaroff's answer, for he gave me what I asked for: a hint to help me in the right direction. –  timvermeulen May 24 '13 at 22:00
@timjver: (I tend to give hints, but in this case I consider the difference between the two answers pretty negligible.) In future if you want only a hint, you should probably say so explicitly: the homework does not necessarily mean that only a hint is appropriate (or wanted), and phrases like help me in the right direction are also used by people who want any amount of help, up to and including a full solution. Given that I’ve seen full solutions offered even when the explicit request was made, you might even want to put it in boldface. –  Brian M. Scott May 24 '13 at 22:07
Right, I'll keep that in mind for future questions. :) Thank you. –  timvermeulen May 24 '13 at 22:09
@timjver: You’re welcome. (I just wish that I’d realized that you really did just want a hint; it’s always a pleasure to see that kind of request.) –  Brian M. Scott May 24 '13 at 22:12