Ellie Castro
2022-04-30
Answered

Please, conclude by the Bounded Convergence Theorem that the sequence is convergent $\ne}^{-n$

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Trey Harrington

Answered 2022-05-01
Author has **10** answers

Let $a}_{n}={\ne}^{-n$

$\Rightarrow {a}_{n}=\frac{n}{{e}^{n}}$

${a}_{n}min=\frac{n}{\mathrm{\infty}}=0$

For maximum

Put $\frac{d}{dn}\left[{a}^{n}\right]=0$

$\frac{d}{dn}\left[\frac{n}{{e}^{n}}\right]=0$

$\frac{{e}^{n}-{\ne}^{n}}{{\left({e}^{n}\right)}^{2}}=0$

${e}^{n}(1-{n}^{1})=0$

${e}^{n}=0$ $n=1$

$\therefore n=1$ ${e}^{n}>0,\mathrm{not}\mathrm{possible}$

$a}_{n}max=\frac{1}{e$

Thus, $0\le {a}_{n}\le \frac{1}{e}$

Given sequence is bounded

$\frac{d}{dn}\left[{a}^{n}\right]>0$

$\Rightarrow {e}^{n}(1-n)>0$

$\Rightarrow 1-n>0$

$\Rightarrow 1>n$

$n<1$

For $n<1$ sequence is decreasing, thus it is convergent

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