To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$. 
Now looking at the series 
\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 &=\sum\limits_{i=1}^\infty \dfrac{\left(-1\right)^{n+1}\,2^n}{n}
\end{align}
How do I relate these two series?

Hint: a common series that is used for computing log of any real number is
$$
\log\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\dots\right)
$$
$u=\frac{1+x}{1-x}\iff x=\frac{u-1}{u+1}$
