Compute $\sum\limits_{n=0}^{\infty}\frac{1}{(2n+1) . 4^n} $

$\displaystyle\sum_{n=0}^{\infty}\frac{1}{(2n+1) . 4^n} $
Can anyone tell me how to approach? It must be convergent as $\displaystyle\sum_{n=0}^{\infty}\frac{1}{4^n}$ is convergent. and answer will be less than 4/3.
but how to approach I can not see?

Solutions Collecting From Web of "Compute $\sum\limits_{n=0}^{\infty}\frac{1}{(2n+1) . 4^n} $"

$$\sum_{n=1}^\infty\dfrac1{(2n+1)4^n}=2\sum_{n=1}^\infty\dfrac{(1/2)^{2n+1}}{2n+1}$$

Now for $1\ge x>-1,$ $$\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots$$

$$\ln(1+x)-\ln(1-x)=?$$

Hint. One may write
$$
\sum_{n=0}^{\infty}\frac{1}{(2n+1)4^n}=2\sum_{n=0}^{\infty}\int_0^{1/2}t^{2n}dt=2\int_0^{1/2}\sum_{n=0}^{\infty}t^{2n}dt=\int_0^{1/2}\frac2{1-t^2}\:dt
$$ the latter may be evaluated by a partial fraction decomposition.

$$\sum_{n=0}^\infty q^n = \frac{1}{1-q}$$
$$\sum_{n=0}^\infty x^{2n} = \frac{1}{1-x^2}$$
$$\int\sum_{n=0}^\infty x^{2n} = \int\frac{1}{1-x^2}$$
$$\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1} = \tanh^{-1}(x)$$
$$\sum_{n=0}^\infty \frac{x^{2n}}{2n+1} = \frac{\tanh^{-1}(x)}{x}$$
$$\sum_{n=0}^\infty \frac{t^{n}}{2n+1} = \frac{\\tanh^{-1}(\sqrt{t})}{\sqrt{t}}$$
Now evaluate at $t=1/4$