# Please, conclude by the Bounded Convergence Theorem that the sequence

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

Let ${a}_{n}={\ne }^{-n}$
$⇒{a}_{n}=\frac{n}{{e}^{n}}$
${a}_{n}min=\frac{n}{\mathrm{\infty }}=0$
For maximum
Put $\frac{d}{dn}\left[{a}^{n}\right]=0$
$\frac{d}{dn}\left[\frac{n}{{e}^{n}}\right]=0$
$\frac{{e}^{n}-{\ne }^{n}}{{\left({e}^{n}\right)}^{2}}=0$
${e}^{n}\left(1-{n}^{1}\right)=0$
${e}^{n}=0$ $n=1$
$\therefore n=1$
${a}_{n}max=\frac{1}{e}$
Thus, $0\le {a}_{n}\le \frac{1}{e}$
Given sequence is bounded
$\frac{d}{dn}\left[{a}^{n}\right]>0$
$⇒{e}^{n}\left(1-n\right)>0$
$⇒1-n>0$
$⇒1>n$
$n<1$
For $n<1$ sequence is decreasing, thus it is convergent