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$\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?

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

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 $t=1/4$

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