study-the-convergence-of-u-n-k-0-n-1-C-n-k-with-C-n-k-n-k-n-k-

Question Number 30194 by abdo imad last updated on 17/Feb/18

s t u d y t h e c o n v e r g e n c e o f u_{n} = \sum_{k = 0}^{n} \frac{1}{C_{n}^{k}} w i t h

C_{n}^{k} = \frac{n!}{k! (n - k)!} .

Commented by prof Abdo imad last updated on 22/Feb/18

l e t p u t x_{n} = \frac{1}{C_{n}^{k}} \forall n \in N x_{n} > 0 w e h a v e

\frac{x_{n + 1}}{x_{n}} = \frac{C_{n}^{k}}{C_{n + 1}^{k}} = \frac{\frac{n!}{k! (n - k)!}}{\frac{(n + 1)!}{k! (n + 1 - k)!}} = \frac{1}{n + 1} \frac{(n + 1 - k)!}{(n - k)!}

= \frac{n + 1 - k}{n + 1} = 1 - \frac{k}{n + 1} < 1 f r o m t h a t u_{n} i s c o n v e r g e n t .