$1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ..........$ $1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ......$
Can anyone help me out how to solve this. 
My try : I was thinking about the expansion of $tan^{-1}x$. But in that series  positive and negative will come alternatively. 

A different approach would be to use binomial theorem . Consider the series expansion of $\ln (\frac {1}{1-x}),\ln (\frac {1}{1+x})$ using binomial expression for negative index we have $\ln (\frac {1}{1-x})-\ln (\frac {1}{1+x})=2 (x+\frac {x^3}{3}+\frac {x^5}{5}+.. ) $ now pull out $x $ common from $RHS $ and put $x=\frac {1}{2} $ thus we have $\frac {1}{2x}\ln(\frac {1}{1-x})-\ln(\frac {1}{1+x})=S $ thus $S=\ln (3) $
