Question

$\frac{1}{\mathrm{1.2.3}}+\frac{5}{\mathrm{3.4.5}}+\frac{9}{\mathrm{5.6.7}}+...\infty =\frac{5}{2}-3{\log }_{e}b2$

Answer 1

Here $R.H.S.=\frac{5}{2}-{\log }_{e}8$
$=\frac{5}{2}-{\log }_e{2}^3$
$=\frac{5}{2}-3{\log }_{e}2$
$=\frac{1}{2}(5-6{\log }_{e}2)$
The above results (B) and (C) are
${\log }_{e}2=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...$ ...(1)
${\log }_{e}2=1-\frac{1}{2.3}-\frac{1}{4.5}-\frac{1}{6.7}+...$ ...(2)
The R.H.S. has number 5 and $6{\log }_{e}2$ in numerator then multiplying (2) by (5) and then adding in (1), we have
$6{\log }_{e}2=5(1-\frac{1}{2.3}-\frac{1}{4.5}-\frac{1}{6.7}+...)+(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...)$
$=5-(\frac{5}{2.3}-\frac{1}{1.2})-(\frac{5}{4.5}-\frac{1}{3.4})-(\frac{5}{6.7}-\frac{1}{5.6})-...$
$6{\log }_{e}2=5-\frac{2}{\mathrm{1.2.3}}-\frac{10}{\mathrm{3.4.5}}-\frac{18}{\mathrm{5.6.7}}+...$
Dividing both sides by -2, then
$-3{\log }_{e}2=-\frac{5}{2}+\frac{1}{\mathrm{1.2.3}}+\frac{5}{\mathrm{3.4.5}}+\frac{9}{\mathrm{5.6.7}}+...$
Hence
$\frac{1}{\mathrm{1.2.3}}+\frac{5}{\mathrm{3.4.5}}+\frac{9}{\mathrm{5.6.7}}+...=\frac{5}{2}-3{\log }_{e}2=\frac{5}{2}-{\log }_{e}8$