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$n\in\mathbb{N}$, find the limit \begin{equation} \lim_{n\to\infty} e^{\frac{n}{4}}n^{-\frac{n+1}{2}}(1^1\cdot2^2\cdot\cdots\cdot n^n)^{\frac{1}{n}} \end{equation} I calculate the limit in the following way:

Express the third term in this \begin{equation} (1^1\cdot2^2\cdot\cdots\cdot n^n)^{\frac{1}{n}}=\left(\frac{(n!)^{n+1}}{1!\cdots n!}\right)^{\frac{1}{n}}=\left(\frac{(n!)^{n+1}}{\prod_{k=1}^{n}k!}\right)^{\frac{1}{n}} \end{equation} Apply Stirlings forluma, the denominator can be approximated as \begin{equation} \prod_{k=1}^{n}k!=\prod_{k=1}^{n}\sqrt{2\pi}k^{k+\frac{1}{2}}e^{-k}=(2\pi)^{\frac{n}{2}}(n!)^{\frac{1}{2}}e^{-\frac{n(n+1)}{2}}\prod_{k=1}^{n}k^{k} \end{equation} Use Euler-Maclaurin's summation formula, the last term tends to \begin{equation} \prod_{k=1}^{n}k^{k}=\exp\left(\sum_{k=1}^{n}k\ln k\right)\sim\exp\left(\frac{1}{2}n^2\ln n-\frac{1}{4}n^2+\frac{1}{2}n\ln n+\frac{1}{12}\ln n+O(1)\right) \sim n^{\frac{n(n+1)}{2}}e^{\frac{n^2}{4}} \end{equation} Substitute it back to the limit and did an annoying elimination, finally got the answer $1$. Is there any other better way to change the third term to an easier form?

Cunyi Nan
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    You could also do this: Let $a_n$ be that expression that you want the limit, raised to $n$. Compute instead, the limit of $a_{n+1}/a_n$. Stolz-Cesaro gives that this is the limit of $a_n^{1/n}$ too, which is the one you want. This way you remove the need to deal with the factorials. You will need to use $\lim(1+1/n)^n=e$ – NDB Jul 08 '23 at 16:14

2 Answers2

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As an alternative we can use that

$$\frac{a_{n+1}}{a_n}=L \implies \sqrt[n]{a_n}=L$$

and by

$$a_n=e^{\frac{n^2}{4}}n^{-\frac{n(n+1)}{2}}(1^1\cdot2^2\cdot\cdots\cdot n^n)$$

we obtain

$$\frac{a_{n+1}}{a_n}=\frac{e^{\frac{n^2+2n+1}{4}}(n+1)^{-\frac{(n+1)(n+2)}{2}}(1^1\cdot2^2\cdot\cdots\cdot n^n\cdot (n+1)^{n+1})}{e^{\frac{n^2}{4}}n^{-\frac{n(n+1)}{2}}(1^1\cdot2^2\cdot\cdots\cdot n^n)}=e^{\frac{2n+1}{4}}\left(\frac{n}{n+1}\right)^{\frac{n(n+1)}{2}}=$$

$$=e^{\frac{2n+1}{4}}e^{\frac{n(n+1)}{2}\log \left(1-\frac{1}{n+1}\right)}=e^{\frac{2n+1}{4}}e^{-\frac{n}{2}-\frac{n}{4(n+1)}+O\left(\frac1n\right)}=e^{O\left(\frac1n\right)}\to e^0=1$$

user
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1

$$\prod_{k=1}^{n}k!=H(n)$$ where $H(n)$ is the hyperfactorial function.

$$H(n)=A\,e^{-\frac{n^2}{4}}\, n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\left(1+\frac{1}{720 n^2}+O\left(\frac{1}{n^4}\right) \right)$$ where $A$ is Glaisher's constant.

Raising your expression to power $n$, we then have

$$\left(e^{\frac n 4} n^{-\frac{n+1}{2}}\right)^n \,H(n)=A \sqrt[12]{n}\left(1+\frac{1}{720 n^2}+O\left(\frac{1}{n^4}\right) \right)$$ and then the result $$1+\frac{12 \log (A)+\log (n)}{12 n}+O\left(\frac{1}{n^2}\right)$$

Claude Leibovici
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