Question

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

• A. e
• B. $2e$
• C. e/2
• D. e/3

Answer 1

Answer: (C)

e/2

Answer 2

$\log_{e}1$

Here$\log_{e}n$

$\log_{e}(1 + n)$

$\log_{e}(1 - n)$.

Trick : The sum of this series upto 4 terms is 1.359 ...... and this is value of e/2 approximately.