Computing $\lim_{n \to \infty} \left[\left(\prod_{i=1}^{n}i!\right)^{1\over n^{2}} (n^{x})\right] $ if exists for certain $x\in\mathbb R$

Question

I came across this problem on my exam and even after three hours of trying I am not able to get through the problem.

I think it can be expressed as the limit of a sum but I am not sure and all my attempts to do the same have failed.

Question:

If the limit $$\lim_{n \to \infty} \left[\left(\prod_{i=1}^{n}i!\right)^{\frac{1}{n^2}} (n^{x})\right]$$ exists and is finite what are the possible values of $x$ and the corresponding values of limit?

I got something like $$e^{\ln{1\over n}\displaystyle\sum_{r=1}^n\left[1-{r-1\over n}\ln\left({r\over n}\right)\right]}$$ but I am having trouble with that r-1 over there.

Can this be expressed as a Riemann sum?

(So recently all the answers I have got seem to be skipping steps I get that all these people are professional in their own fields but can you please try to write answers for someone who has considerably low IQ than yourself, thanks for the same)

Answer 1

Asymptotic Expansion via Riemann Sum

Compute the log of the product as a Riemann Sum $$ \begin{align} \frac1{n^2}\sum_{k=1}^n k\log(n-k+1) &=\sum_{k=1}^n\frac{k}{n}\left(\log\left(1-\frac{k}{n}+\frac1n\right)+\log(n)\right)\frac1n\tag{1a}\\ &\sim\int_0^1x\log(1-x)\,\mathrm{d}x+\frac12\log(n)\tag{1b}\\ &=\int_0^1\log(1-x)\,\mathrm{d}\frac{x^2-1}2+\frac12\log(n)\tag{1c}\\ &=-\int_0^1\frac{x+1}2\,\mathrm{d}x+\frac12\log(n)\tag{1d}\\ &=\frac12\log(n)-\frac34\tag{1e} \end{align} $$ Thus, the product is asymptotically $$ \left(\prod_{k=1}^nk!\right)^{1/n^2}\sim e^{-3/4}n^{1/2}\tag2 $$ Therefore, for $x=-1/2$, the limit comes to $$ \bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\left(\prod_{k=1}^nk!\right)^{1/n^2}n^{-1/2}=e^{-3/4}}\tag3 $$ For $x\lt-1/2$, the limit is $0$.

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Asymptotic Expansion via Euler-Maclaurin Sum Formula

As is shown in this answer, we have asymptotically in $n$, $$ \sum_{k=1}^n k^{-z} =\zeta(z)+\frac{n^{1-z}}{1-z}+\frac12n^{-z}-\frac{z}{12}n^{-1-z}+O\left(\frac1{n^{3+z}}\right)\tag4 $$ applying $-\frac{\mathrm{d}}{\mathrm{d}z}$: $$ \begin{align} \sum_{k=1}^n\log(k)k^{-z} &=-\zeta'(z)+n^{1-z}\frac{(1-z)\log(n)-1}{(1-z)^2}+\frac12\log(n)n^{-z}\\ &-n^{-1-z}\frac{z\log(n)-1}{12}+O\!\left(\frac{\log(n)}{n^{3+z}}\right)\tag5 \end{align} $$ Setting $z=0$: $$ \sum_{k=1}^n\log(k)=\overbrace{\,\,-\zeta'(0)\ }^{\frac12\log(2\pi)}+n(\log(n)-1)+\frac12\log(n)+\frac1{12n}+O\!\left(\frac{\log(n)}{n^3}\right)\tag6 $$ Setting $z=-1$: $$ \begin{align} \sum_{k=1}^n\log(k)k &=\overbrace{-\zeta'(-1)}^{\log(A)-\frac1{12}}+n^2\frac{2\log(n)-1}4+\frac12n\log(n)+\frac{\log(n)+1}{12}\\ &+O\!\left(\frac{\log(n)}{n^2}\right)\tag7 \end{align} $$ where $A$ is the Glaisher–Kinkelin constant.

Thus, $$ \begin{align} \sum_{k=1}^n(n-k+1)\log(k) &=n^2\frac{2\log(n)-3}4+n\log\left(\frac{\sqrt{2\pi}}en\right)+\frac5{12}\log(n)\\ &+\log\left(\frac{\sqrt{2\pi}}{A}\right)+\frac1{12}+\frac1{12n}+O\!\left(\frac{\log(n)}{n^2}\right)\tag8 \end{align} $$ and therefore, $$ \prod_{k=1}^nk!=\frac{\sqrt{2\pi}}{A}e^{1/12}\,\color{#C00}{n^{n^2/2}e^{-3n^2/4}}\color{#090}{\left(\frac{\sqrt{2\pi}}en\right)^n}n^{5/12}\color{#00F}{e^{\frac1{12n}+O\left(\frac{\log(n)}{n^2}\right)}}\tag9 $$ Finally, $$ \bbox[5px,border:2px solid #C0A000]{\left(\prod_{k=1}^nk!\right)^{1/n^2}=\color{#C00}{n^{1/2}e^{-3/4}}+\color{#090}{O\!\left(\frac{\log(n)}{n^{1/2}}\right)}}\tag{10} $$

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The Glaisher–Kinkelin Constant

Equation $(6)$ is essentially Stirling's Formula: $$ \prod_{k=1}^nk=\sqrt{2\pi}\,n^{n+1/2}e^{-n}\left(1+\frac1{12n}+O\!\left(\frac1{n^2}\right)\right)\tag{11} $$ where $\sqrt{2\pi}=e^{-\zeta'(0)}$. This shows that $\zeta'(0)=-\frac12\log(2\pi)$.

Equation $(7)$ says that $$ \prod_{k=1}^nk^k=A\,n^{n^2/2+n/2+1/12}e^{-n^2/4}\left(1+O\!\left(\frac{\log(n)}{n^2}\right)\right)\tag{12} $$ where $A=e^{\frac1{12}-\zeta'(-1)}$.

Just as $$ \sum_{k=1}^n\frac1k=\log(n)+\gamma+O\!\left(\frac1n\right)\tag{13} $$ is the defining limit for $\gamma$, the Euler-Mascheroni Constant, $(12)$ appears to be the defining limit for $A$, the Glaisher–Kinkelin constant.

Answer 2

Here is a more accurate than needed estimate from my answer at Formula for $1! \times 2! \times \cdots \times n!$?:

$\prod\limits_{k=0}^{n} n! \sim C^{1/2} (2\pi)^{3/8}n^{5/12}(2\pi)^{n/2}(n/e)^n \left(\dfrac{n}{e^{3/2}}\right)^{n^2/2} $ where $C =\lim\limits_{n \to \infty} \dfrac1{n^{1/12}}\prod\limits_{k=1}^n\left( \dfrac{k!}{\sqrt{2\pi k}(k/e)^k} \right) \approx 1.046335066770503 $.

Taking the $n^2$ root, $\left(\prod\limits_{k=0}^{n} n!\right)^{1/n^2} \to\left(\dfrac{n}{e^{3/2}}\right)^{1/2} $ since all the other terms go to $1$.

Answer 3

If the Riemann sum is convergent, that means that the left Riemann sum limit is the same as the right Riemann sum limit. That means $$\lim_{n\to \infty}\frac1n\sum_{r=1}^n \frac{r-1}n\ln\frac{r-1}n=\lim_{n\to \infty}\frac1n\sum_{r=1}^n \frac{r}n\ln\frac{r}n$$ Now just use the squeeze theorem, since each term in $\frac{r-1}{n}\ln\frac rn$ is between the corresponding terms in the left and right Riemann sums.

Answer 4

If we take logarithm of sequence in question we get another sequence $$a_n=\frac{1}{n^2}\left(\sum_{k=1}^{n}\log k! +xn^2\log n\right)\tag{1}$$ Applying Cesaro-Stolz we can see that limit of $a_n$ is same as that of $$\frac{1} {2n-1}\left(\log n! +xn^2\log n-x(n-1)^2\log n-x(n-1)^2\log(1-1/n)\right) $$ provided the limit of above exists. The above expression has the same limit as that of $$\frac{1}{2n-1}(\log n! +2nx\log n)+\frac{x}{2}\tag{2}$$

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We can now use Stirling approximation which says that $$\log n! - \left(n\log n - n+\frac{1}{2}\log(2\pi n) \right) \to 0$$ which implies that $$\frac{1}{2n-1}(\log n! - n\log n+n) \to 0$$ Now expression $(2)$ can be rewritten as $$\frac{\log n! - n\log n+n} {2n-1}+(1+2x)\frac{n\log n}{2n-1}-\frac{n}{2n-1}+\frac {x} {2}$$ Clearly the first term tends to $0$ and the last two terms have a finite limit $(x-1)/2$. The second term has a finite limit only when $x=-1/2$. If $x<-1/2$ the expression in $(2)$ tends to $-\infty $ and if $x>-1/2$ the expression tends to $\infty$.

It follows that the limit in question exists if $x\leq - 1/2$. If $x<-1/2$ then the desired limit is $0$ and if $x=-1/2$ then the desired limit is $e^{(x-1)/2}=e^{-3/4}$.

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Another alternative is to apply Cesaro-Stolz on first term in $(2)$. Doing this gives us the expression $$\frac{1}{2}\left(\log n+2xn\log n-2x(n-1)\log n-2x(n-1)\log(1-1/n)\right)$$ which has same limit as that of $$\frac{1+2x}{2}\log n+x$$ and the conclusion is same as obtained earlier. This approach avoids complicated Stirling formula.