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Determine, with proof, the value of the limit $L = \lim\limits_{n\to\infty} \dfrac{2^{n^{k}}}{n!}$ where $k=1.1.$

I think it's infinite because $\frac{2^{n^k}}{n!}\ge (\frac{2^{n^{0.1}}}{n})^n=(2^{n^{0.1}-\log n})^n\to \infty.$

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    $\begingroup$ Sometimes our intuition leads us astray, which is fine.... Hint: $n! \le n^n$ and $\frac{2^{n^{1.1}}}{n^n} = \big( \frac{2^{n^{0.1}}}n \big)^n$. $\endgroup$ Aug 15, 2021 at 17:49
  • $\begingroup$ Write down what a(n+1) / a(n) is. For large n it becomes large. $\endgroup$
    – gnasher729
    Aug 15, 2021 at 18:03

3 Answers 3

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"I'm pretty sure the limit with $k=1.1$ should evaluate to $0$" Well, no ...

Let $a_n= \dfrac{2^{n^{1.1}}}{n!}.$ Then

$$\tag1 \ln a_n = n^{1.1}\ln 2- \ln n! \ge n^{1.1}\ln 2- n\ln n = n(n^{.1}\ln 2-\ln n).$$

Since $n^{.1}\ln 2-\ln n \to \infty,$ we see $(1)\to \infty.$ It follows that $a_n\to \infty.$

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Define $a_n:=\frac{2^{n^k}}{n!}$. I will show that $$\lim_{n\rightarrow\infty}a_n=\left\{\begin{matrix} \infty &\text{if} & k>1\\ 0 & \text{if} & k\leq 1 \end{matrix}\right. $$

  • If $k>1$, then by Bernoulli's inequality $$(n+1)^k-n^k=n^k\Big(\big(1+\frac{1}{n}\big)^k-1\big)\geq n^k\frac{k}{n}=n^{k-1}k$$ Hence $$ a_{n+1}=\frac{1}{n+1}\frac{2^{n^k}}{n!}2^{(n+1)^k-n^k}=\frac{2^{(n+1)^k-n^k}}{n+1}a_n\geq \frac{(2^k)^{n^{k-1}}}{n+1}a_n$$ Recall that for any $a>1$ and $\beta>0$ $$\frac{x+1}{{a}^{x^\beta}}=\frac{\big(x^{\beta}\big)^{\tfrac{1}{\beta}}+1}{{a}^{x^\beta}}\stackrel{y=x^{\beta}}{=}\frac{y^{1/\beta}+1}{a^y}\xrightarrow{y\rightarrow\infty}0$$ Thus, for $k>1$, there is $n_0$ such that for $n\geq n_0$ $a_{n+1}>2a_n$. This means that $$a_{n+n_0}\geq 2^n a_{n_0}\xrightarrow{n\rightarrow\infty}\infty$$

  • For $k\leq 1$, $n^k\leq n$ and so, $$\frac{2^{n^k}}{n!}\leq \frac{2^n}{n!}\xrightarrow{n\rightarrow\infty}0$$

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By ratio test, for any $k>1$

$$\dfrac{2^{(n+1)^{k}}}{(n+1)!}\dfrac{n!}{2^{n^{k}}}=\frac{2^{(n+1)^{k}-n^k}}{n+1}\ge \frac{2^{kn^{k-1}}}{n+1}\to \infty$$

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