TITLE: Find the sum of the series $1+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{5}\cdot\frac{1}{4^2}+\frac{1}{7}\cdot\frac{1}{4^3}+\cdots$
QUESTION [6 upvotes]: Find the sum of the series : $$1+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{5}\cdot\frac{1}{4^2}+\frac{1}{7}\cdot\frac{1}{4^3}+\cdots$$

REPLY [9 votes]: This can be transformed to $$\sum_{n=1}^{\infty} \frac{2}{(2n-1)2^{2n-1}}$$
Let $$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)}$$
Then, we have $f'(x)=\sum_{n=1}^{\infty} x^{2n-2} = \frac{1}{1-x^2}$.
Therefore, we have $$f(x)=\int \frac{1}{1-x^2} = \frac{1}{2} \ln \frac{x+1}{1-x}+C$$
It is clear that $C=0$.
Now plugging $x=\frac{1}{2}$ in this equation, we have $\sum_{n=1}^{\infty} \frac{1}{(2n-1)2^{2n-1}} = \frac{1}{2} \ln 3$, so the desired answer is double that number, or $\ln 3$.