calculate-n-2-k-2-1-k-n-k-

Question Number 61804 by maxmathsup by imad last updated on 09/Jun/19

c a l c u l a t e \sum_{n = 2}^{\infty} \sum_{k = 2}^{\infty} \frac{1}{k^{n} k!}

Commented by maxmathsup by imad last updated on 15/Jun/19

S = \sum_{k = 2}^{\infty} \frac{1}{k!} \sum_{n = 2}^{\infty} {(\frac{1}{k})}^{n} = \sum_{k = 2}^{\infty} \frac{w_{k}}{k!}

w_{k} = \sum_{n = 2}^{\infty} {(\frac{1}{k})}^{n} = \sum_{n = 0}^{\infty} {(\frac{1}{k})}^{n} - 1 - \frac{1}{k} = \frac{1}{1 - \frac{1}{k}} - 1 - \frac{1}{k} = \frac{k}{k - 1} - 1 - \frac{1}{k}

= \frac{k - 1 + 1}{k - 1} - 1 - \frac{1}{k} = \frac{1}{k - 1} - \frac{1}{k} \Rightarrow S = \sum_{k = 2}^{\infty} \frac{1}{k!} (\frac{1}{k - 1} - \frac{1}{k})

= \sum_{k = 2}^{\infty} \frac{1}{(k - 1) k!} - \sum_{k = 2}^{\infty} \frac{1}{k k!}

= \sum_{k = 1}^{\infty} \frac{1}{k (k + 1)!} - \sum_{k = 2}^{\infty} \frac{1}{k k!} l e t f i n d \sum_{k = 2}^{\infty} \frac{1}{k k!} = \sum_{k = 1}^{\infty} \frac{1}{k k!} - 1

w e h a v e e^{x} = \sum_{n = 0}^{\infty} \frac{x^{k}}{k!} \Rightarrow \int_{0}^{x} e^{t} d t = \sum_{k = 0}^{\infty} {[\frac{t^{k + 1}}{(k + 1) k!}]}_{0}^{x} + c \Rightarrow

e^{x} - 1 = \sum_{k = 0}^{\infty} \frac{x^{k + 1}}{(k + 1) k!}

x = 1 \Rightarrow e - 1 = \sum_{k = 0}^{\infty} \frac{1}{(k + 1) k!}

\sum_{k = 1}^{\infty} \frac{1}{k (k + 1)!} = \sum_{k = 1}^{\infty} \frac{1}{k (k + 1) k!} = \sum_{k = 1}^{\infty} (\frac{1}{k} - \frac{1}{k + 1}) \frac{1}{k!} = \sum_{k = 1}^{\infty} \frac{1}{k k!} - \sum_{k = 1}^{\infty} \frac{1}{(k + 1) k!}

\Rightarrow S = \sum_{k = 1}^{\infty} \frac{1}{k k!} - \sum_{k = 1}^{\infty} \frac{1}{(k + 1) k!} - \sum_{k = 2}^{\infty} \frac{1}{k k!}

= 1 - {\sum_{k = 0}^{\infty} \frac{1}{(k + 1) k!} - 1} = 1 - {e - 1 - 1} = 1 - e + 2 = 3 - e \Rightarrow

S = 3 - e .

Answered by perlman last updated on 09/Jun/19

\sum_{n = 2}^{\infty} \sum_{k = 2}^{\infty} \frac{1}{k^{n} k!} = \sum_{k = 2}^{+ \infty} \sum_{n = 2}^{+ \infty} \frac{1}{k^{n} k!}

j u s t i f y

\forall k, n ⩾ 2

\sum_{n = 2}^{+ \infty} \sum_{k = 2}^{+ \infty} \frac{1}{k! k^{n}} < \sum_{n} \frac{e}{2^{n}} = e

\sum_{n = 2}^{\infty} \sum_{k = 2}^{\infty} \frac{1}{k^{n} k!} = \sum_{k = 2}^{+ \infty} \sum_{n = 2}^{+ \infty} \frac{1}{k^{n} k!} = \sum_{k} \frac{1}{k!} \sum_{n = 2} \frac{1}{k^{n}}

= \sum_{k} \frac{1}{k!} . \frac{1}{k^{2}} . \frac{k}{k - 1} = \sum_{k ⩾ 2} \frac{1}{k! k (k - 1)} = \sum_{k} \frac{k - (k - 1)}{k! k (k - 1)} = \sum_{k} \frac{1}{k! (k - 1)} - \frac{1}{k! k}

= \sum_{k} \frac{k - (k - 1)}{k! (k - 1)} - \sum_{k} \frac{1}{k k!}

= \sum_{k} \frac{1}{(k - 1)! (k - 1)} - \sum_{k = 2}^{+ \infty} \frac{1}{k!} - Σ \frac{1}{k! k}

= \sum_{k = 2}^{+ \infty} \frac{1}{(k - 1)! (k - 1)} - \sum_{k = 2}^{+ \infty} \frac{1}{k! k} - \sum_{k = 0}^{+ \infty} \frac{1}{k!} + \sum_{k = 0}^{1} \frac{1}{k!}

= \frac{1}{(2 - 1)! (2 - 1)} - e + 1 + 1 = 3 - e

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