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.
On
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)
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.