Show that $\sum_{n=0}^\infty\frac{1}{(2n+1)3^{2n+1}}=$ ln$(\sqrt2)$

Question

Question: Show that $\sum_{n=0}^\infty\frac{1}{(2n+1)3^{2n+1}}=$ ln$(\sqrt2)$

My attempt: I attempted a partial fractions decomposition of the general term and could not produce anything meaningful. I'm thinking log laws or the taylor series for ln$(x)$ might be useful here but I'm not sure what to do.

Would love if someone could point me in the right direction. Thanks.

Answer 1

You have for $|x| <1$:

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots=\sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k}$$

$$-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots=\sum_{k=1}^\infty \frac{x^k}{k}$$

so by summing: $$\ln(1+x)-\ln(1-x)=2 x+0+2 \frac{x^3}{3}+\cdots=2 \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}$$ i.e: $$\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}=\ln\left(\sqrt{\frac{1+x}{1-x}}\right)$$ with $x=\frac{1}{3}$ you obtain: $$\ln \left(\sqrt{\frac{4\div 3}{2\div 3}}\right)=\ln(\sqrt{2})$$