Question

${\log }_{e}a-\frac{1}{b}=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{3.4.5}}+\frac{1}{\mathrm{5.6.7}}+...\infty$. Find $a+b^2$.

Answer 1

Here $L.H.S.={\log }_{e}2-\frac{1}{2}$
$\frac{2{\log }_{e}2-1}{2}$
and we know that
${\log }_{e}2=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...$ ...(1)
and ${\log }_{e}2=1-\frac{1}{2.3}-\frac{1}{4.5}-\frac{1}{6.7}-...$ ...(2)
the L.H.S. has number 1 and $2{\log }_{e}2$ in numerator, then adding (1) and (2), we have
$2{\log }_{e}2=1+(\frac{1}{1.2}-\frac{1}{2.3})+(\frac{1}{3.4}-\frac{1}{4.5})+(\frac{1}{5.6}-\frac{1}{6.7})+...$
or $2{\log }_{e}2-1=\frac{2}{\mathrm{1.2.3}}+\frac{2}{\mathrm{3.4.5}}+\frac{2}{\mathrm{5.6.7}}+...$
Dividing both sides by 2, we get
${\log }_{e}2-\frac{1}{2}=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{3.4.5}}+\frac{1}{\mathrm{5.6.7}}+...\infty$
Proved.