Summation of this series ${1\over 2}\left[1+{1 \over 3}\left({1 \over 4}\right)^2+ {1 \over 5}\left({1 \over 4}\right)^4+\cdots\right]$

Hint:

$$\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}=\frac{1}{x}\int_{0}^{x}\left(\sum_{n=0}^{\infty}u^{2n}\right)du$$

Can you take it from here?

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Some more explanations:

Notice that each term in your sum can be written like $$\frac{x^{2n}}{2n+1}\tag{1}$$ where $x=1/4$. So your sum can be written: $$\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{2n}}{2n+1}\tag{2}$$Now we know that the anti-derivative of $x^{2n}$ is $\frac{x^{2n+1}}{2n+1}$, and this looks very much like $\frac{x^{2n}}{2n+1}$, except that there is an $x$ too much!

We can fix this problem by taking the anti-derivative, and then divide by $x$. This would look like $$\frac{x^{2n}}{2n+1}=\frac{1}{x}\int_{0}^{x}u^{2n}du=\frac{1}{x}\left(\frac{x^{2n+1}}{2n+1}-\frac{0^{2n+1}}{2n+1}\right)\tag{3}$$ for each term. Now you can just take the sum of both sides, and you should get your original sum.

You can now write you sum as an integral, with integrand $$\sum_{n=0}^{\infty}u^{2n}\tag{4}$$ can you simplify this?

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Just to finish:

You end up with the integral in the hint, and the sum $(4)$ is a geometric series which converges to $$\frac{1}{1-u^2}=\frac{1}{2}\left(\frac{1}{1-u}+\frac{1}{1+u}\right)\tag{5}$$ We can now compute the integral: $$\frac{1}{4x}\int_{0}^{x}\frac{1}{1-u}+\frac{1}{1+u}du=\frac{1}{4x}\ln\left(\frac{1+x}{1-x}\right)\tag{6}$$ Setting $x=1/4$ yields: $$\ln\left(\frac{5}{3}\right)\approx 0.51$$

Hint:You may use following function:

$$S=1+\frac{1}{3}x^2+\frac{1}{5}x^4+ . . . $$

Multiplying both sides by x we get:

$$S.x=x+\frac{1}{3}x^3+\frac{1}{5}x^5+ . . .$$

Taking derivative we get:

$$ S'x+ S=1+x^2+x^4 +. . . (x^2)^n=\frac{(x^2)^{n+1} -1}{x^2-1}$$

Taking integral we get:

$$S.x=\int \frac{(x^2)^{n+1} -1}{x^2-1}$$

$$S=\frac{1}{x}\int \frac{(x^2)^{n+1} -1}{x^2-1}$$

Put $x=\frac {1}{4}$.