Answer
$\dfrac{3}{2} \ln (2)=1+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{4}+.....$
Work Step by Step
We have $\ln (2)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.... ~~~~~(a)$
and $\dfrac{1}{2}\ln (2)=\dfrac{1}{2}(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+....) ~~~~~(b)$
or, $\dfrac{1}{2}\ln (2)=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{8}+.... ~~~~~(c)$
Next, we will add equations (b) and (c) to obtain:
$\dfrac{3}{2} \ln (2)=1+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{4}+.....$
