Question

The sum of series $\frac{1}{1.2.3}+\frac{1}{3.4.5}+\frac{1}{5.6.7}+\dots .$is

• A. ${\log }_{e}2–\frac{1}{2}$
• B. ${\log }_{e}2$
• C. ${\log }_{e}2+\frac{1}{2}$
• D. ${\log }_{e}2+1$

Answer 1

The correct option is A

${\log }_{e}2–\frac{1}{2}$

Explanation for the correct option:

Find the required sum.

Given: series $\frac{1}{1.2.3}+\frac{1}{3.4.5}+\frac{1}{5.6.7}+\dots .$

Let$T_n$​ be the $n^{th}$ term of given series.

$$\begin{gathered} \therefore T_n=\frac{1}{[1+(n-1\left)2\right][2+(n-1\left)2\right][3+(n-1\left)2\right]} \\ =\frac{1}{(2n-1)2n(2n+1)} \end{gathered}$$
Let ${\displaystyle \frac{1}{\left(2n-1\right)\left(2n\right)\left(2n+1\right)}}=\left({\displaystyle \frac{1}{2}}\right)\left[{\displaystyle \frac{A}{2n-1}}+{\displaystyle \frac{B}{n}}+{\displaystyle \frac{C}{2n+1}}\right]$ from this we will have

$A\left(2n+1\right)n+B\left(2n+1\right)\left(2n-1\right)+Cn\left(2n-1\right)=1$ by simplifying this equation we get,

$A=1$ , $B=-1$ and $C=1$ .
Now we will have ${\displaystyle \frac{1}{2}}\left\{{\displaystyle \frac{1}{2n-1}}-{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{2n+1}}\right\}$
Let the summation of the series is $S$ .
So we can say that,

$2S=\left({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{1}}+{\displaystyle \frac{1}{3}}\right)+\left({\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{5}}\right)+\left({\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{7}}\right)+.......$
$\Rightarrow 2S=(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{2}{7}+.......)$
$\Rightarrow 1+2S=2(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+..............)$ .
From the basic concepts, we know that ${\log }_{e}x=\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{{\left(-1\right)}^{n-1}{\left(x-1\right)}^n}{n}}$ .
By using this we will have ${\log }_{e}2=1-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{6}}+..............$ .
Therefore we can say that $1+2S=2{\log }_{e}2$

$\Rightarrow$ $S={\log }_{e}2-{\displaystyle \frac{1}{2}}$ .
So, the sum of the given series ${\displaystyle \frac{1}{\mathrm{1.2.3}}}+{\displaystyle \frac{1}{\mathrm{3.4.5}}}+{\displaystyle \frac{1}{\mathrm{5.6.7}}}+....$ is ${\log }_{e}2-{\displaystyle \frac{1}{2}}$.

Hence, option (A) is the correct answer.