# 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?

## 2 Answers

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:

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