Question

The sum of the series $ \frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+....+\infty $ =

• A. $log(2e)$
• B. $log(e/2)$
• C. $log(4/e)$
• D. $None\, of \,these$

Answer 1

The Correct Option is B

$\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\ldots+\infty$ Here, $T_{n}=\frac{1}{2n.\left(2n+1\right)}$ $T_{n}=\frac{1}{2n}-\frac{1}{2n+1}$ $T_{n}=\frac{1}{2n}-\frac{1}{2n+1}$ $T_{1}=\frac{1}{2}-\frac{1}{3}$ $T_{2}=\frac{1}{4}-\frac{1}{5}$ $T_{3}=\frac{1}{6}-\frac{1}{7}\ldots\infty$ $\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$ $\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$ Adding all, $\left(T_{1}+T_{2}+T_{3}+\ldots+T_{\infty}\right)$ $=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\right)-\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots\right)\ldots\left(i\right)$ We know that, $log\,2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots\infty$ From E $\left(i\right)$, $\left(T_{1}+T_{2}+\ldots+T_{\infty}\right)=-\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots\infty\right)$ $=-\left\{\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots\infty\right)-1\right\}$ $=-\left\{log_{e}\,2-log_{e} e\right\}$ $=-log\left(2/ e\right)$ $=log \left(e/ 2\right)$