Evaluate $ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots$ Evaluate $$ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots$$
All i could do was to see that $$\frac{1}{3}=\frac{1}{2.1+1},\frac{1}{5}=\frac{1}{2.2+1},\frac{1}{7}=\frac{1}{2.3+1},\dots$$
SO, we should be able to write 
$$ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots=1+\sum_{n=1}^{\infty}\frac{1}{2n+1}.\frac{1}{4^n}$$
This is what i could simplify the given question to... 
But, this is doing me no good.. I could not go any further..
I would be thankful if some one can help me to clear this.
I would request users who are trying to help me to just give necessary hints but not post it as answer.
Thank You.

From this, for $|x|<1,$
$$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$$
$$\text{and }\ln(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4+\cdots$$
Now subtract
Observe that  $$ 1 + \frac13\frac14+\frac15\frac1{4^2}+\frac17\frac1{4^3}+\cdots$$
$$=2\sum_{0\le r<\infty}\frac1{2r+1}\left(\frac12\right)^{2r+1}$$
