# Compute $\lim_{n \to +\infty} n^{-\frac12 \left(1+\frac{1}{n}\right)} \left(1^1 \cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\frac{1}{n^2}}$

How to compute $$\displaystyle \lim_{n \to +\infty} n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(1^1\cdot 2^2 \cdot 3^3 \cdots n^n \right)^{\dfrac{1}{n^2}}$$ I'm interested in more ways of computing limit for this expression

My proof:

Let $u_n$be that sequence we've:

\begin{eqnarray*} \ln u_n &=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln k\\ &=& -\frac{n+1}{2n}\ln n + \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}+\frac{1}{n^2}\sum_{k=1}^n k\ln n\\ &=& \frac{1}{n^2}\sum_{k=1}^n k\ln \frac{k}{n}\\ &=& \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\ln \frac{k}{n}\\ &\to&\int_0^1 x\ln x\,dx = -1/4 \end{eqnarray*}

Therefore the limit is $e^{-\frac{1}{4}}$

• Look good to me – DeepSea Jan 24 '16 at 9:54
• This can also followed straighforwardly from the asymptotics of the hyperfactorial which i derived here: math.stackexchange.com/questions/1571739/… – tired Jan 24 '16 at 22:33

Another approach, considering $$A_n= n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} \left(\prod_{i=1}^n i^i \right)^{\dfrac{1}{n^2}}=n^{-\dfrac12 \left(1+\dfrac{1}{n}\right)} H(n)^{\dfrac{1}{n^2}}$$ where appears the hyperfactorial function. Taking logarithms $$\log(A_n)={-\dfrac12 \left(1+\dfrac{1}{n}\right)}\log(n)+\dfrac{1}{n^2}\log(H(n))$$ Now, using Stirling like approximations $$\log(H(n))=n^2 \left(\frac{1}{2} \log(n)-\frac{1}{4}\right)+\frac{1}{2} n \log (n))+\left(\log (A)+\frac{1}{12} \log \left(n\right)\right)+O\left(\frac{1}{n^2}\right)$$ where appears Glaisher constant.
All of that, once recombined, leads to $$\log(A_n)=-\frac{1}{4}+\frac{\log (A)+\frac{1}{12} \log \left(n\right)}{n^2}+O\left(\frac{1}{n^{5/2}}\right)$$ which shows the limit and how it is approached.
We have $$\exp \left (- \frac {1} {2} \left (1 + \frac {1} {n}\right) \log n + \frac {1} {n^2} (1 \log 1 + 2 \log 2 + \cdots + n \log n)\right) = \exp \left (- \frac {1} {2} \left (1 + \frac {1} {n}\right) \log n + \frac {1} {2} \log n - \frac {1} {4} + \frac {\log n} {2n} + o \left (\frac {1} {n}\right) \right) = \exp \left (-\frac {1} {4} + o \left (\frac {1} {n}\right) \right),$$ using the fact that $$1 \log 1 + 2 \log 2 + \cdots + n \log n = \frac {1} {2} n^2 \log n - \frac {1} {4} n^2 + \frac {1} {2} n \log n + o (n).$$ By letting $n \to \infty$, we have $$\exp \left (-\frac {1} {4} + o \left (\frac {1} {n}\right) \right) \to \exp (-\frac {1} {4}).$$