How to find the limit of $\frac{(-3)^n}{n!}$ when $n$ goes to $\infty$ by the squeeze theorem or any other method?
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1Hint: when you increase $n$ by $1$, you multiply the term by $-3/n$. – Oct 12 '15 at 07:24
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1Possible duplicate of Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. – Oct 12 '15 at 15:09
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$\sum_{n=1}^\infty \frac{(-3)^n}{n!}$ converges because of the ratio test: $\lim_{n\to\infty}\left\lvert\frac{(-3)^{n+1}/(n+1)!}{(-3)^n/n!} \right\rvert = \lim_{n\to\infty}\left\lvert\frac{-3}{n+1}\right\rvert = 0 <1$. But this means that $\lim_{n\to\infty}\frac{(-3)^n}{n!} = 0$.
Claudius
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HINT:
$3^n\lt n!$ for sufficiently large $n$ and $\lim (-1)^n$ does not exist.
Bumblebee
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Hint: For $n > 6$ we have that $\left|\frac{(-3)^n}{n!}\right| < \left|\frac{(-3)^6}{6!}\left(\frac{-1}{2}\right)^{n-6}\right|$. Now it's bounded by a sequence that you can show converges to $0$.
Luke
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