Un = Σ_(k=1) ^n (1/ ((n),(k) )) show that the sequence converges and determine the limit


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 201070 by Rodier97 last updated on 29/Nov/23

$U n = \underset{k = 1}{\sum^{n}} \frac{1}{(\begin{matrix} n \\ k \end{matrix})}$ $s h o w t h a t t h e s e q u e n c e c o n v e r g e s a n d$ $d e t e r m i n e t h e l i m i t$
	Answered by Frix last updated on 29/Nov/23

	$\frac{1}{(\begin{matrix} n \\ k \end{matrix})} = \frac{k! (n - k)!}{n!}$ $U_{n} = 1 + \underset{k = 1}{\sum^{n - 1}} \frac{k! (n - k)!}{n!} = 1 + \frac{1}{n!} \underset{k = 1}{\sum^{n - 1}} k! (n - k)!$ $The greatest summands are those with$ $k = 1, k = n - 1 . Their sum is \frac{2}{n} . The next$ $are k = 2, k = n - 2 . Their sum is \frac{4}{n^{2} - n}$ $k = 3, k = n - 3 \Rightarrow \frac{12}{n^{3} - 3 n^{2} + 2 n}$ $. . .$ $\frac{2}{n}$ $\frac{2}{n} + \frac{4}{n^{2} - n} = \frac{2 n + 2}{n^{2} - n}$ $\frac{2}{n} + \frac{4}{n^{2} - n} + \frac{12}{n^{3} - 3 n^{2} + 2 n} = \frac{2 n^{2} - 2 n + 8}{n^{3} - 3 n^{2} + 2 n}$ $. . .$ $We end up with \frac{{a n}^{p} + . . .}{n^{p + 1} + . . .}$ $\lim_{n \to \infty} \frac{{a n}^{p} + . . .}{n^{p + 1} + . . .} = 0$ $\Rightarrow$ $\lim_{n \to \infty} U_{n} = 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum