What is the sum of series $1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$ I am trying to solve this series question which is 
$$1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$$
Till now I've only been able to write the general form of this series which is 
$$\sum_{n=1}^\infty\frac1{(2n-1)4^{n-1}}$$
but I don't know how to proceed. 
 A: Hint: Consider the series
$$\ln(1+x)-\ln(1-x)$$
for $x=\dfrac12$.
A: \begin{align}
& \sum_{k=1}^\infty\frac1{(2k-1)4^{k-1}} \\
=& \sum_{k=0}^\infty\frac1{(2k+1)4^{k}} \\
=& \sum_{k=0}^\infty  4^{-k} \int_0^1 x^{2k} dx \\
=& \int_0^1 \sum_{k=0}^\infty  \left(\frac{x^2}{4}\right)^k dx \tag{Fubini-Tonelli Theorem} \\
=& \int_0^1 \frac{1}{1-x^2/4} dx \\
=& 2 \int_0^{1/2} \frac{1}{1-t^2} dt \tag{$t = x/2$} \\
=& \int_0^{1/2} \left( \frac{1}{1-t} + \frac{1}{1+t} \right) dt \tag{$t = x/2$} \\
=& \left[ -\ln|1-t| + \ln|1+t| \right]_0^{1/2} \\
=& -\ln(1/2) + \ln(3/2) \\
=& \ln 3
\end{align}
A: Hint Reindexing your expression for the series for convenience gives
$$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k .$$
Now, the factors $\frac{1}{2 k + 1}$ suggest considering the power series
$$\operatorname{artanh} x \sim \sum_{k = 0}^\infty \frac{1}{2 k + 1} x^{2 k + 1} .$$

Rewrite the given series to get $$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k = 2 \sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\color{red}{\frac{1}{2}}\right)^{2 k + 1} = 2 \operatorname{artanh} \color{red}{\frac{1}{2}} = 2 \cdot \frac{1}{2}\log \frac{1 + \color{red}{\frac{1}{2}}}{1 - \color{red}{\frac{1}{2}}} = \boxed{\log 3} .$$

