# Why $\sum_{k=0}^\infty (-1)^k \frac{(\ln 4)^k}{k!}=0.25$?

Hi can someone please explain why

$$\sum_{k=0}^{\infty} \frac{(-1)^k (\ln 4)^k}{k!} = 0.25$$

• Do you know the Taylor expansion of $e^x$? – lulu Aug 26 '15 at 13:03
• Hint: $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$ – robjohn Aug 26 '15 at 13:05
That expression is same as $e^{-ln(4)}$, and this is in fact $1/4$, i.e $0.25$
First note that $$\exp(x)=\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}$$ So now $$\sum\limits_{k=0}^{\infty} \frac{(-1)^k (\ln 4)^k}{k!} =\sum\limits_{k=0}^{\infty} \frac{(-\ln 4)^k}{k!}$$ $$=\sum\limits_{k=0}^{\infty} \frac{(\ln 4^{(-1)})^k}{k!} =\sum\limits_{k=0}^{\infty} \frac{\left(\ln \frac14\right)^k}{k!}$$ $$=\exp\left(\ln \frac14\right)=\frac14$$
Hint: what is $e^x$ as a series? Can you express $0.25$ in terms of an exponential base $e$?