Find sum of the series $\sum\limits_{n=0}^{\infty}\frac{1}{2n+1}\left ( \frac{1}{2} \right )^{n}$

Since$$\sum_{n=0}^\infty x^{2n}=\sum_{n=0}^\infty(x^2)^n=\frac1{1-x^2}$$and since$$\left(\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}\right)'=\sum_{n=0}^\infty x^{2n}=\frac1{1-x^2}\text{ and }\sum_{n=0}^\infty\frac{0^{2n+1}}{2n+1}=0,$$you have$$\sum_{n=0}^\infty\frac{x^{2n+1}}{2n+1}=\int_0^x\frac{\mathrm dt}{1-t^2}=\operatorname{arctanh}x.$$Therefore$$x\neq0\implies\sum_{n=0}^\infty\frac{x^{2n}}{2n+1}=\frac{\operatorname{arctanh}x}x.$$In particular\begin{align}\sum_{n=0}^\infty\frac1{2n+1}\left(\frac12\right)^n&=\sum_{n=0}^\infty\frac1{2n+1}\left(\frac1{\sqrt2}\right)^{2n}\\&=\frac{\operatorname{arctanh}\left(\frac1{\sqrt2}\right)}{\frac1{\sqrt2}}\\&=\sqrt 2\operatorname{arctanh}\left(\frac1{\sqrt2}\right)\\&=\frac1{\sqrt2}\log\left(\frac{\sqrt2+1}{\sqrt2-1}\right).\end{align}

\begin{align} & \sum_{n=0}^\infty \frac{(1/2)^n}{2n+1} = \sum_{n=0}^\infty \frac{(1/\sqrt2\,)^{2n}}{2n+1} \\[8pt] = {} & \sqrt 2 \sum_{n=0}^\infty \frac{(1/\sqrt2\,)^{2n+1}}{2n+1} \\[10pt] & \frac d {dx} \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1} = \sum_{n=0}^\infty x^{2n} = \frac 1 {1-x^2}. \end{align} So you need an antiderivative of $$ \frac 1 {1-x^2} = \frac 1 {(1-x)(1+x)} = \frac A {1-x} + \frac B {1-x}. $$ You need to complete the partial-fractions problem. To find the value of the constant of integration, use the fact that when $x=0$ in the sum above, then the sum is $0.$