Question

$\huge \sf \red{\sum \limits_{n = 2}^{ \infty} \sum \limits_{k = 2}^{ \infty} \frac{1}{ {k}^{n} \cdot k!} }$​

Answer 1

Answer:

$\boxed{ \huge \purple{3 - e}}$

Step-by-step explanation:

Switch the summation signs

$s = \sum \limits_{k= 2}^{ \infty} \sum \limits_{n = 2}^{ \infty} \frac{1}{ {k}^{n} \cdot k!} \\ \\ = \sum \limits_{k= 2}^{ \infty} \frac{1}{k!} .\sum \limits_{n = 2}^{ \infty} \frac{1}{ {k}^{n} }$

First sum is a infinite geometric series

$= \sum \limits_{k= 2}^{ \infty} \frac{1}{k!} . \frac{( \frac{1}{ k {}^{2} }){ } }{1 - \frac{1}{k} } \\ \\ = \sum \limits_{k= 2}^{ \infty} \frac{1}{k!} . \frac{1}{ {k}^{2} - k } \\ \\ = \sum \limits_{k= 2}^{ \infty} \frac{1}{k!} . \frac{k - (k - 1)}{k(k - 1)} \\ \\ = \sum \limits_{k= 2}^{ \infty} \frac{1}{k!} . \bigg( \frac{1}{k - 1} - \frac{1}{k} \bigg ) \\ \\ \sum \limits_{k= 2}^{ \infty} \bigg( \frac{1}{(k - 1)k!} - \frac{1}{k.k!} \bigg ) \\ \\ = \frac{1}{1.2!} - \frac{1}{2.2!} + \frac{1}{2.3!} - \frac{1}{3.3!} + \frac{1}{3.4!} - .... \\ \\ = \frac{1}{2} - \bigg( \frac{1}{2.2!} - \frac{1}{2.3!} \bigg) - \bigg( \frac{1}{3.3!} - \frac{1}{3.4!} \bigg) - .... \\ \\ = \frac{1}{2} - \bigg( \frac{3 - 1}{2.3!} \bigg) - \bigg( \frac{4 - 1}{3.4!} \bigg) - .... \\ \\ = \frac{1}{2} - \bigg( \frac{2}{2.3!} \bigg) - \bigg( \frac{3}{3.4!} \bigg) - .... \\ \\ = \frac{1}{2} - \frac{1}{3!} - \frac{1}{4!} - \frac{1}{5!} - .... \\ \\ = \frac{1}{2} - \bigg( \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + .... \bigg)$

We know that

[ In expansion formula put x = 1 ]

$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} +.. \\ \\ \implies \: \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} +.. = e - 1 - 1 - \frac{1}{2} \\ \\ \implies \: \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + .... = e - \frac{5}{2}$

Put this value in sum

$s = \frac{1}{2} - (e - \frac{5}{2} ) \\ \\ \implies \: s = \frac{1}{2} - e + \frac{5}{2} \\ \\ \implies \: s = 3 - e$

Answer 2

$\huge{Answer : }$

$\huge\red{s=3-e}$