Un-k-1-n-1-n-k-show-that-the-sequence-converges-and-determine-the-limit-

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}

\dots

\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}

\dots

We end up with \frac{{a n}^{p} + \dots}{n^{p + 1} + \dots}

\lim_{n \to \infty} \frac{{a n}^{p} + \dots}{n^{p + 1} + \dots} = 0

\Rightarrow

\lim_{n \to \infty} U_{n} = 1

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