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  
Consider the power series
$$
f(x) := \sum\limits_{n=1}^\infty \dfrac{x^n}{2n-1}
$$
Investigate what differential equation it satisfies.  Looking at it suggests that this is
maybe easier for
$$
g(x) := \frac{f(x^2)}{x} = \sum\limits_{n=1}^\infty \dfrac{x^{2n-1}}{2n-1}
$$
