Question

Compute the sum upto infinite terms .

$\sf{ 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + ....... \: \infty }$ ​

Answer 1

Question:-

$1 + \frac{1}{2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + .....infinite \\ 1 + \frac{1}{2} + ( \frac{1}{2} - \frac{1}{3} ) + ( \frac{1}{3} - \frac{1}{4})+ ..... + \frac{1}{(infinity - 1) - infinity} \\ analysing \: the \: terms \: we \: </strong><strong>can</strong><strong> \: say \: that \\ all \: the \: terms \: will \: be \: cancelled \: out \: \\ except \: 1 </strong><strong>,</strong><strong>\:</strong><strong>1</strong><strong>/</strong><strong>2</strong><strong>,</strong><strong>1</strong><strong>/</strong><strong>2</strong><strong> and \: \frac{1}{infinity} \\ \frac{1}{infinity} = 0 \\ so \: final \: value = 1</strong><strong>+</strong><strong>1</strong><strong>/</strong><strong>2</strong><strong>+</strong><strong>1</strong><strong>/</strong><strong>2</strong><strong>$=2

Answer 2

Answer:

1/infinity=0

=> 1+1/2+1/6+1/12

=> 12+6+2+1/12

=> 21/12 ans.