k-1-i-0-k-1-n-i-k-is-this-really-mean-something-

Question Number 37859 by kunal1234523 last updated on 18/Jun/18

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

is this really mean something

Commented by math khazana by abdo last updated on 18/Jun/18

\prod_{i = 0}^{k - 1} (n - i) = n (n - 1) (n - 2) \dots (n - k + 1)

= \frac{n (n - 1) \dots . (n - k + 1) (n - k) (n - k - 1 \dots . 2.1}{(n - k)!}

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

\sum_{k = 1}^{\infty} (\dots) = \sum_{k = 1}^{\infty} C_{n}^{k} a n d t h i s i s a n o n s e n s

b e c a u s e k ? m u s t b e ⩽ n b y d e f i n i t i o n o f

c o m b i n a t i o n \dots