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Find the limit of the sequence $$\frac{c^n}{n!^{\frac{1}{k}}}$$, $(k>0, c>0)$

Now when $0<c<1$ we get $$0<\frac{c^n}{n!^{\frac{1}{k}}}< \frac{1}{n!^{\frac{1}{k}}}$$ So by Sandwich Theorem we get the limit of the sequenc equal to $0$ But when $c>1$ i do not understand how to move?

user677491
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2 Answers2

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Let $k>0$, $c>0$ and $$ \forall n \in \mathbb{N}, \, u_n =\frac{c^n}{n!^{\frac 1 k}} $$ We have $$ \lim\limits_{n \to +\infty}\frac{u_{n+1}}{u_n} = \lim\limits_{n \to + \infty}\frac{c}{(n+1)^{\frac 1 k}} = 0 $$ So by the ratio test $$ \lim\limits_{n \to +\infty}u_{n} = 0 $$

Monadologie
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Since everything is positive, $L$ is that limit if and only if $L^k=\lim_{n\to\infty}\frac{c^n}{n!}$. Now, for all $h\ge M=\left\lceil c\right\rceil+1$, $\frac{c}{h}\le \frac cM<1$, so that for all $n\ge M$ $$0\le \frac{c^n}{n!}\le \frac{c^{M-1}}{(M-1)!}\left(\frac{c}{M}\right)^{n-M+1}\to 0$$

Because $\frac cM<1$ and $\frac{c^{M-1}}{(M-1)!}\cdot\frac{c^{1-M}}{M^{1-M}}=\frac{M^{M-1}}{(M-1)!}$ is a constant. So $L^k=0$.