I need some help with this limit
$$\lim_{n\to \infty} \left(\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}}\right)^{1/n^2}$$
This problem appears in a chapter about Riemann sums, so I think I must split the fractions in the parantheses:
$$\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}} = \frac{1^1}{n^1}\times \frac{2^2}{n^2}\times ... \times \frac{n^n}{n^n}$$
Now I am stuck.
Hint. By taking the logarithm we find $$\log\left(\left(\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}}\right)^{1/n^2}\right)=\frac{1}{n}\sum_{k=1}^n (k/n)\log(k/n)$$ which looks like a Riemann sum...
Let $I$ denote the given limit. Then $$I = \lim_{n\to\infty}\left(\prod_{i=1}^n\left(\dfrac{i}{n}\right)^i\right)^{\frac{1}{n^2}}$$ $$=e^{\lim_{n\to\infty}\frac{1}{n^2}\sum_{i=1}^n\left(i\log\frac{i}{n}\right)}$$ $$ = e^{\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\left(\frac{i}{n}\log\frac{i}{n}\right)}$$ $$=e^{\int_0^1x\log x\mathrm{d}x}$$ Now apply integration by parts.
$$\left(\frac{1^{1}\cdot2^{2}\cdot\cdot\cdot n^{n}}{n^{\left(1+2+...+n\right)}}\right)^{\frac{1}{n^{2}}}=\exp\left(\frac{1}{n^{2}}\ln\left(\frac{1^{1}\cdot2^{2}\cdot\cdot\cdot n^{n}}{n^{\left(1+2+...+n\right)}}\right)\right)$$$$=\exp\left(\frac{1}{n^{2}}\ln\left(\prod_{k=1}^{n}\left(\frac{k}{n}\right)^{k}\right)\right)=\exp\left(\frac{1}{n}\sum_{k=1}^{n}\ln\left(\left(\frac{k}{n}\right)^{\frac{k}{n}}\right)\right)$$$$=\exp\left(\int_{0}^{1}\ln\left(\left(x\right)^{x}\right)dx\right)=\exp\left(\int_{0}^{1}x\ln\left(\left(x\right)\right)dx\right)$$$$=\exp\left(\frac{x^{2}}{2}\ln\left(\left(x\right)\right)\Big|_0^1-\frac{1}{2}\int_{0}^{1}xdx\right)$$$$=\color{red}{\exp\left(\frac{-1}{4}\right)\simeq0.778800783071}$$
$$A=\lim_{n\to\infty}\left(\prod_{r=1}^n\left(\dfrac rn\right)^{r/n}\right)^{1/n}$$
$$\ln A=\lim_{n\to\infty}\dfrac1n f\left(\dfrac rn\right)$$ where $f(x)=x\ln x$
Now use The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$ and integrate by parts using LIATE : How does it work?
Just for your curiosity
It is clear that Riemann sum make the problem quite simple.
But we can go beyond the limit and approximate the term which is $$a_n=\left(n^{-\frac{1}{2} n (n+1)} H(n)\right)^{\frac{1}{n^2}}\implies \log(a_n)=\frac 1{n^2}\log\left(n^{-\frac{1}{2} n (n+1)} H(n)\right)$$ where $H(n)$ is the hyperfactorial function.
Expanding the logarithm and using the asymptotics of $\log(H(n))$ we quickly end with $$\log(a_n)=\frac{1}{ n^2}\left(-\frac{n^2}{4}+\log (A)+\frac{1}{12} \log \left({n}\right)+\frac{1}{720 n^2}+O\left(\frac{1}{n^4}\right)\right)$$ where $A$ is Glaisher constant that is to say $$\log(a_n)=-\frac{1}{4}+\frac{12\log (A)+ \log \left({n}\right)}{n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it ia approached.
Moreover, this gives you a shortcut method for computing $a_n$. If you are patient, compute $a_{10}$; the exact value is $0.78224014$ while the above (very) truncated series would give $0.78224004$.
