How do we compute this sum?

Question

Given: $$f(x)=\sum_{n=0}^{\infty} \frac{2x^{2n+1}}{2n+1}$$ How do we show that: $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)4^n} = ln3$$

Hints given are that

$$f(1/2)=\sum_{n=0}^{\infty} \frac{1}{(2n+1)4^n}$$

and $$f'(x)=\frac{1}{1-x^2}$$

From what I understand, there isn't a series definition of logarithms is there?

Answer 1

Hint: By integrating $\frac{1}{1 + x}$, $\frac{1}{1 - x}$ together with their respective geometric series, show that $f(x)$ is exactly the taylor expansion of $$\ln\left(\frac{1 + x}{1 - x}\right)$$

and then compute the required series using the fact that it is equal to $f(x)$ at $x = \frac{1}{2}$.

Answer 2

We have: $$ \frac{1}{4^n (2n+1)} = 2\int_{0}^{1/2}x^{2n}\,dx $$ hence: $$ \sum_{n\geq 0}\frac{1}{4^n(2n+1)} = 2\int_{0}^{1/2}\frac{dx}{1-x^2}=\left.\log\left(\frac{1+x}{1-x}\right)\right|_{0}^{1/2}=\color{red}{\log 3}. $$

Answer 3

From $f(x)=\sum_{n=0}^{\infty} \frac{2x^{2n+1}}{2n+1}$

you get $$ f'(x)=\sum_{n=0}^{\infty} {2x^{2n}} = \frac 2{1 - x^2} = \frac 1{1-x} + \frac 1{1+x} $$ So as $f(0) = 0$: $$ f(x) = \log (1+x) - \log(1-x) = \log\frac{1+x}{1-x} $$

From this, just evaluate $f(1/2)$.