$\frac{5}{1·2·3}+\frac{7}{3·4·5}+\frac{9}{5·6·7}+...$ is equal to

The correct option is A

$\log \left(\frac{8}{e}\right)$

Explanation for the correct option:

Step 1: Find the $n$th term of the series:

Given: $\frac{5}{1·2·3}+\frac{7}{3·4·5}+\frac{9}{5·6·7}+...$

It can also be written as $\sum _1^{\infty }\frac{2n+3}{(2n-1)2n(2n+1)}$.

Step 2: Break the obtained expression into partial expressions:

Simplify the expression by breaking it into smaller fractions.

$\frac{2n+3}{(2n-1)2n(2n+1)}=\frac{A}{\left(2n-1\right)}+\frac{B}{2n}+\frac{C}{(2n+1)}$

$\Rightarrow$$\frac{2n+3}{(2n-1)2n(2n+1)}=\frac{2n(2n+1)A+B\left(2n-1\right)(2n+1)+2n(2n-1)C}{\left(2n-1\right)2n(2n+1)}$

$\Rightarrow$ $2n+3=2n(2n+1)A+B\left(2n-1\right)(2n+1)+2n(2n-1)C$

$\Rightarrow$ $2n+3=4n^{2}A+2nA+4n^{2}B-B+4n^{2}C-2nC$

Comparing coefficient of $n^2,n$ and constant

$4n^2\left(A+B+C\right)=0n^2$, $-B=3$, $2n(A-C)=2n$

$\Rightarrow$$A=2,B=-3,C=1$

$\begin{array}{rcl}\therefore \sum _1^{\infty }\frac{2n+3}{(2n-1)2n(2n+1)}& =& \sum _1^{\infty }\left(\frac{2}{2n-1}-\frac{3}{2n}+\frac{1}{2n+1}\right)\\ & =& \sum _1^{\infty }\left(\frac{2}{2n-1}-\frac{2}{2n}\right)+\sum _1^{\infty }\left(\frac{1}{2n+1}-\frac{1}{2n}\right)\\ & =& 2\sum _1^{\infty }\left(\frac{1}{2n-1}-\frac{1}{2n}\right)+\sum _1^{\infty }\left(\frac{1}{2n+1}-\frac{1}{2n}\right)\\ & =& 2\sum _1^{\infty }\left(\frac{1}{2n\left(2n-1\right)}\right)+\sum _1^{\infty }\left(\frac{1}{2n\left(2n+1\right)}\right)\end{array}$

Step 3: Use logarithmic equations to get the summation.

$\begin{array}{rcl}2\sum _1^{\infty }\left(\frac{1}{2n\left(2n-1\right)}\right)+\sum _1^{\infty }\left(\frac{1}{2n\left(2n+1\right)}\right)& =& 2\left(1-\frac{1}{2}+\frac{1}{3}....\right)+\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}-\frac{1}{4}\right)\\ & =& 2\log 2+(\log 2-1)\\ & =& 3\log 2-1\\ & =& \log 8-\log e\\ & =& \mathit{l}\mathit{o}\mathit{g}\left(\frac{8}{e}\right)\end{array}$ $\left[\mathbf{\because }\mathit{l}\mathit{o}\mathit{g}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{=}\mathit{x}\mathbf{-}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{x}}^{\mathbf{3}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\right]$

Hence, option (A) is the correct answer.