Prove $ \sum_{n=1}^\infty \frac{(-1)^n }{2n\cdot4^n} = \ln(2/5^{0.5})$

Question

I proved that the following summation converges but how to prove that the it is equal to $\ln(2/5^{0.5})$

$$ \sum_{n=1}^\infty \frac{(-1)^n }{2n\cdot4^n}$$

Answer 1

Hint: derive $\sum \frac{x^n}{2n}$ calculate the sum, and re-integrate for $x=-\frac 14$.

Answer 2

Use $\ln(1+x) =\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}x^n}{n} $ and put $x = \dfrac14$.

This gives $\ln(1+\frac14) =\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{4^nn} $ so

$\begin{array}\\ \sum_{n=1}^{\infty} \dfrac{(-1)^{n}}{4^n2n} &=-\dfrac12 \ln(5/4)\\ &=-\ln(\sqrt{5/4})\\ &=-\ln(\sqrt{5}/2)\\ &=\ln(2/\sqrt{5})\\ \end{array} $