Question

$\frac{1}{1.2.3} + \frac{1}{3.4.5} + \frac{1}{5.6.7} + .....\infty =$

• A. $\log_{e}\sqrt{2}$
• B. $\log_{e}2 - \frac{1}{2}$
• C. $\log_{e}2$
• D. $\log_{e}4$

Answer 1

Answer: (B)

$\log_{e}2 - \frac{1}{2}$

Answer 2

We know

$e^{\left( x - \frac{1}{2}(x - 1)^{2} + \frac{1}{3}(x - 1)^{3} - \frac{1}{4}(x - 1)^{4} + ....... \right)}$ …..(i)

$\log x$........ …..(ii)

Again, $\log(x - 1)$

$y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ......\infty$ …..(iii)

Add (ii) and (iii),

$x =$

$\log_{e}y$ …..(iv)

$\log_{e}\frac{1}{y}$ .

Aliter : $e^{y}$

$e^{- y}$

$1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + .......\infty =$

Putting $e/2$

$e$,

$2e$

.............................................

.............................................

$3e$

Adding all terms, we get

$\frac{1.2}{1!} + \frac{2.3}{2!} + \frac{3.4}{3!} + \frac{4.5}{4!} + .....\infty = 2e$

$3e$

$3e - 1$

$e$.