let-u-n-k-1-n-1-k-n-k-calculate-n-1-u-n-

Question Number 45957 by maxmathsup by imad last updated on 19/Oct/18

l e t u_{n} = \sum_{k = 1}^{n} \frac{1}{k! (n - k)!} c a l c u l a t e \sum_{n = 1}^{\infty} u_{n}

Commented by maxmathsup by imad last updated on 20/Oct/18

w e h a v e u_{n} = \frac{1}{n!} \sum_{k = 1}^{n} \frac{n!}{k! (n - k)!} = \frac{1}{n!} \sum_{k = 1}^{n} C_{n}^{k}

= \frac{1}{n!} (\sum_{k = 0}^{n} C_{n}^{k} - 1) = \frac{1}{n!} {2^{n} - 1} = \frac{2^{n}}{n!} - \frac{1}{n!} \Rightarrow

\sum_{n = 1}^{\infty} u_{n} = \sum_{n = 1}^{\infty} \frac{2^{n}}{n!} - \sum_{n = 1}^{\infty} \frac{1}{n!} = \sum_{n = 0}^{\infty} \frac{2^{n}}{n!} - 1 - \sum_{n = 0}^{\infty} \frac{1}{n!} + 1

= e^{2} - e .

Answered by Smail last updated on 19/Oct/18

u_{n} = \underset{k = 1}{\sum^{n}} \frac{1}{k! (n - k)!} = \underset{k = 1}{\sum^{n}} \frac{n!}{k! (n - k)!} \times \frac{1}{n!}

= \frac{1}{n!} \underset{k = 1}{\sum^{n}} \frac{n!}{k! (n - k)!} = \frac{1}{n!} (\underset{k = 0}{\sum^{n}} \frac{n!}{k! (n - k)!} - 1)

{l e t}^{'} s t a k e p (x) = {(1 + x)}^{n} = \underset{k = 0}{\sum^{n}} \frac{n!}{k! (n - k)!} x^{k}

S o p (1) = 2^{n} = \underset{k = 0}{\sum^{n}} \frac{n!}{k! (n - k)!}

u_{n} = \frac{1}{n!} (2^{n} - 1)

\underset{n = 1}{\sum^{\infty}} u_{n} = \underset{n = 1}{\sum^{\infty}} \frac{2^{n}}{n!} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n!}

e^{x} = \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!}

e^{1} = \underset{n = 0}{\sum^{\infty}} \frac{1}{n!}

e^{2} = \underset{n = 0}{\sum^{\infty}} \frac{2^{n}}{n!}

S o \underset{n = 1}{\sum^{\infty}} u_{n} = \underset{n = 0}{\sum^{\infty}} \frac{2^{n}}{n!} - 1 - (\underset{n = 0}{\sum^{\infty}} \frac{1}{n!} - 1)

= e^{2} - e

Commented by maxmathsup by imad last updated on 20/Oct/18

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