<s>[INST] $1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ..........$ $1 + \frac{1}{3}\frac{1}{4} + \frac{1}{5}\frac{1}{4^2} + \frac{1}{7}\frac{1}{4^3} + ......$
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. 
 [/INST] $$\sum_{r=0}^\infty\dfrac1{(2r+1)}\left(\dfrac12\right)^{2r}=2\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)}$$ where $2x=1$
Now $-\ln(1-x)=\sum_{n=0}^\infty\dfrac{x^n}n$ for $-1\le x<1$
$\ln(1+x)-\ln(1-x)=?$
Alternatively, $$\sum_{r=0}^\infty\dfrac{x^{2r+1}}{(2r+1)}=\int\sum_{r=0}^\infty(x^2)^rdx$$
</s>