Evaluating the series: $1 +(1/3)(1/4) +(1/5)(1/4^2)+(1/7)(1/4^3)+ \cdots$ 
Sum the series:
  $$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots$$

How can I solve this? I am totally stuck on this problem.
 A: Let $$S = \sum_{n=0}^{\infty} \dfrac1{2n+1} \dfrac1{4^n}$$
Note that
$$\dfrac1{2n+1} = \int_0^1 x^{2n} dx$$
Hence, we can write
\begin{align}
S & = \sum_{n=0}^{\infty} \dfrac1{4^n}\int_0^1 x^{2n} dx = \int_0^1 \sum_{n=0}^{\infty} \left(\dfrac{x^2}{4} \right)^n dx\\
& = \int_0^1 \dfrac{dx}{1-\dfrac{x^2}4} = \int_0^1 \dfrac{dx}{2-x} + \int_0^1 \dfrac{dx}{2+x}\\
& = \left. -\log(2-x) \right \vert_0^1 + \left. \log(2+x) \right \vert_0^1 = \log 3
\end{align}
Hence,
$$\sum_{n=0}^{\infty} \dfrac1{2n+1} \dfrac1{4^n} = \log(3)$$
A: Using Infinite geometric series,  $$1+x+x^2+\cdots=\frac1{1-x}$$ for $|x|<1$
Applying integration wrt to $x, \log(1-x)=-x-\frac{x^2}2-\frac{x^3}3+\cdots+c$ where $c$ is an arbitrary constant of indefinite integral.
Putting $x=0,\log(1)=c\implies c=0$
So, $$\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3+\cdots$$
Putting $x=y$ we get, $$\log(1-y)=-y-\frac{y^2}2-\frac{y^3}3+\cdots$$
Putting $x=-y$ we get, $$\log(1+y)=y-\frac{y^2}2+\frac{y^3}3+\cdots$$
So, $$\log(1+y)-\log(1-y)=2(y+\frac{y^3}3+\frac{y^5}5+\cdots)$$ for $|y|<1$
Consequently, $$1 + \dfrac13 \cdot \dfrac14 + \dfrac15 \cdot \dfrac1{4^2} + \dfrac17 \cdot \dfrac1{4^3} + \cdots=2\left(\frac12+\frac{\left(\frac12\right)^3}3+\frac{\left(\frac12\right)^5}5+\cdots\right)$$
$$=\log(1+\frac12)-\log(1-\frac12)=\log\left(\frac{\frac32}{\frac12}\right)=\log 3$$
