While computing the dimension of certain Specht modules I encountered following limit:
$$\lim\limits_{n \rightarrow \infty} (2^2. 3^3 \ldots n^n)^{\frac{2}{n(n+1)}}$$
In case there would be written $n!$ instead of $2^2. 3^3 \ldots n^n$ one could use Stirling formula in order to compute the limit, but this limit I do not see how to compute it. Intuitively I would except it to converge to $\infty$.
Do somebody has an idea? In any case thanks!
$$\ln((2^2\cdot 3^3\cdots n^n)^{\frac{2}{n(n+1)}})=2\frac{2\ln 2+3\ln 3+\cdots +n\ln n}{n(n+1)}\\\lim_{n\to\infty}2\frac{2\ln 2+3\ln 3+\cdots+n\ln n}{n(n+1)}=\lim_{n\to\infty}2\frac{(n+1)\ln(n+1)}{(n+2)(n+1)-(n+1)n}=\lim_{n\to\infty} 2\frac{\ln(n+1)}{2}=\infty$$