limit of binomial sequence

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Source: Moldova 2000 Grade 12 P2

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#1 h Apr 27, 2021, 8:44 PM • 1 Y

Y by Svyatoslav

For $n\in\mathbb N$ , define
$a_n=\frac1{\binom n1}+\frac1{\binom n2}+\ldots+\frac1{\binom nn}.$ (a) Prove that the sequence $b_n=a_n^n$ is convergent and determine the limit.
(b) Show that $\lim_{n\to\infty}b_n>\left(\frac32\right)^{\sqrt3+\sqrt2}$ .

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#2 h Apr 27, 2021, 11:45 PM

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The answer to the first part of your question is $e^2$ and the second part of your question is silly and so I'm not gonna answer that.

To show that $\lim_{n\to\infty}b_n=e^2,$ we use this general result that if $\{x_n\}$ is a sequence of real number with $x_n > -1$ and $\lim_{n\to\infty}nx_n=x,$ then $\lim_{n\to\infty}(1+x_n)^n=e^x.$ Now let
$x_n:=\sum_{k=1}^{n-1}\frac{1}{\binom{n}{k}}.$ Then $b_n=a_n^n=(1+x_n)^n$ and
$\lim_{n\to\infty}nx_n=\lim_{n\to\infty}n\sum_{k=1}^{n-1}\frac{1}{\binom{n}{k}}=2+\lim_{n\to\infty}n\sum_{k=3}^{n-3}\frac{1}{\binom{n}{k}}. \ \ \ \ \ \ \ \ \ (*)$ We also have
$0 < n\sum_{k=3}^{n-3}\frac{1}{\binom{n}{k}} < \frac{n(n-5)}{\binom{n}{3}}$ and so, by the squeeze theorem, $\lim_{n\to\infty}n\sum_{k=3}^{n-3}\frac{1}{\binom{n}{k}}=0.$ Thus $\lim_{n\to\infty}nx_n=2,$ by $(*),$ and the result follows.

This post has been edited 1 time. Last edited by ysharifi, Apr 28, 2021, 12:12 AM
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