A series involving $\prod_1^n k^k$ Is this series $$\sum_{n\geq 1}\left(\prod_{k=1}^{n}k^k\right)^{\!-\frac{4}{n^2}} $$ convergent or divergent?
My attempt was to use the comparison test, but I'm stuck at finding the behaviour of $\displaystyle \prod_1^n k^k$ as $n$ goes to infinity. Thanks in advance.
 A: By Cesaro-Stolz theorem, we have that $$\lim_{n\to\infty}\frac{1}{n^2 \log(n)}\sum_{k=1}^{n} k\log(k)=\frac{1}{2}$$ and then we see   that for $n$ large, we have that $$\frac{1}{n^2}\sum_{k=1}^{n} k\log(k)\approx\frac{\log(n)}{2}\Rightarrow \frac{-4}{n^2}\sum_{k=1}^{n} k\log(k)\approx -2\log(n)$$and thus $$\left(\prod_{k=1}^{n}k^k\right)^{\!-\frac{4}{n^2}}\approx \frac{1}{n^2}$$
whence we conclude the series converges.
Q.E.D. (an elementary way , without integrals)
A: Hint:
$$\log\left(\prod_{k=1}^n k^k\right) = \sum_{k=1}^n k\log k\cdots $$
A: Here is an elementary approach, without using the Glaisher-Kinkelin constant.
Observe that
$$
\ln \left(\left(\prod_{k=1}^{n}k^k\right)^{\!-\frac{4}{n^2}}\right)= -\frac{4}{n^2} \sum_{k=1}^{n}k\ln k.
$$
Let $k\geq 1$ and let $x \in [k,k+1]$. Since $\displaystyle x \rightarrow x\ln x$ is an increasing function, you may write
$$
k\ln k \leq x\ln x \leq (k+1)\ln (k+1), 
$$ integrating
$$
\int_k^{k+1} k\ln k \:dx \leq \int_k^{k+1}x\ln x \:dx \leq \int_k^{k+1}(k+1)\ln (k+1) \:dx
$$
equivalently,
$$
 \int_{k-1}^{k}x\ln x \:dx \leq k\ln k \leq \int_k^{k+1}x\ln x \:dx 
$$
then summing from $k=1$ to $n$, $n\geq1$, you get
$$
 \int_{0}^{n}x\ln x \:dx \leq \sum_{k=1}^{n}k\ln k \leq \int_1^{n+1}x\ln x \:dx. 
$$
As $n$ tends to $+\infty$, you readily have
$$
\begin{align}
& \int_{0}^{n}x\ln x \:dx  = -\frac{n^2}{4}  \left(1+ \ln \frac{1}{n^2}\right),\\
& \int_{1}^{n+1}x\ln x \:dx  =  -\frac{n^2}{4} \left(1+ \ln \frac{1}{n^2}\right)+\mathcal{O}{(n \ln n)},
\end{align}
$$
giving
$$
1+ \ln \frac{1}{n^2}-\mathcal{O}{\left(\frac{\ln n}{n}\right)} \leq -\frac{4}{n^2} \sum_{k=1}^{n}k\ln k \leq 1+ \ln \frac{1}{n^2} 
$$
thus, by exponentiation, 
$$
\left(\prod_{k=1}^{n}k^k\right)^{\!-\frac{4}{n^2}} \sim \frac{e}{n^2}, \quad n \rightarrow \infty,
$$
and the given series is convergent.
