Question

The sum of the infinite series $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \frac{4}{5!} + ......$is.

• A. $e - 2$
• B. $\frac{2}{3}e - 1$
• C. 1
• D. 3/2

Answer 1

Answer: (C)

1

Answer 2

$\log_{a}b$

= $\log_{e}a - \log_{e}b$

$\log_{e}a + \log_{e}b$

= $\frac{1}{5} + \frac{1}{2}.\frac{1}{5^{2}} + \frac{1}{3}.\frac{1}{5^{3}} + .....\infty =$

= $\log_{e}\frac{\sqrt{5}}{2}$

= $2\log_{e}\frac{\sqrt{5}}{2}$ $\log_{e}{}\lbrack(1 + x)^{1 + x}(1 - x)^{1 - x}\rbrack =$

= $\frac{x^{2}}{2} + \frac{x^{4}}{4} + \frac{x^{6}}{6} + ....\infty$ = 1.