$$\lim_{n\to \infty}\sqrt[n]{\frac{10n!}{n!^{10}}}$$
So we can solve $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}$
$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{10(n+1)}{(n+1)^{10}}=0$$
Which is incorrect, where did I get it wrong?
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1You can't use the ratio test since it's only used to test convergence or divergence. – Toby Mak Nov 28 '18 at 22:32
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1The limit is, in fact, 0: https://www.wolframalpha.com/input/?i=lim_(n-%3Einf)((10n!)%2F(n!)%5E10)%5E(1%2Fn)) – Dr. Mathva Nov 28 '18 at 22:38
2 Answers
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We have that
$$\frac{a_{n+1}}{a_n}=\frac{10(n+1)!}{(n+1)!^{10}}\frac{n!^{10}}{10n!}=\frac{(n+1)}{(n+1)^{10}}\to 0$$
then using "ratio-root" criterion
$$\sqrt[n] {a_n} \to 0$$
user
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More generally, to get $\lim_{n\to \infty}\sqrt[n]{\dfrac{(an)!}{n!^{a}}} $, use the lazy person's "stirling" approximation $\dfrac{(n!)^{1/n}}{n} \to \dfrac1{e} $.
Then, noting that the "$a$" multiplies $n$ before taking the factorial,
$\begin{array}\\ \sqrt[n]{\dfrac{(an)!}{n!^{a}}} &=\dfrac{(an)!^{1/n}}{n!^{a/n}}\\ &=\dfrac{((an)!^{1/(an)})^{a}}{(n!^{1/n})^a}\\ &\to\dfrac{(an/e)^{a}}{(n/e)^a}\\ &\to a^a\\ \end{array} $
When $a=10$ the result is $10^{10}$.
Note that the denominator inside the root is $(10n)!$, not $10(n!)$. In that case, the limit is zero.
marty cohen
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