Proving bounds on an infinite product 
Let $p$ be an infinite product, such that $p = 2^{1/4}3^{1/9}4^{1/16}5^{1/25} ...$ 
Prove that $2.488472296 ≤ p ≤ 2.633367180$.

I start this problem by representing p in the infinite product notation:
 $$p = \displaystyle \prod_{k=2}^{\infty}k^{1/k^2} $$
I also defined the partial product as $p_n = \displaystyle \prod_{k=2}^{n}k^{1/k^2} $
Taking the logarithm:
$p_n = e^{\ln(p_n)} = e^{\sum_{k=2}^{n} \ln(k^{1/k^2})}$
I have no idea how to prove it. Any help would be highly appreciated.
 A: Let $f(x) = \frac{\ln(x)}{x^2}$, where $x \in \mathbb{R}$
Since $f$ is continuous, decreasing, and positive for all values of $x >3$. Then, we can apply the Remainder Estimate for Integral Test. Hence, by definition we have:
$$\int_{n+1}^{\infty} f(x)dx + S_n \leq S \leq \int_n^{\infty}f(x)dx + S_n \  \ \ (*)$$
Note that $P_n = e^{S_n}$, where $S_n = \sum_{k=2}^n \frac{\ln k}{k^2}$
Therefore, since $e$ is always increasing function, we can multiply $e$ to $ (*)$. Thus we obtain:
$$\normalsize e^{\int_{n+1}^{\infty} f(x)dx + S_n} \leq P \leq e^{\int_n^{\infty}f(x)dx + S_n} $$
Consider $n =5$. With an aid of computer software, we have that:
 $2.488472296≤p≤2.633367180$.
A: Here's a thought.  Note $$\log p_n = \sum_{k=2}^n \frac{\log k}{k^2}.$$  Interestingly, Mathematica gives the closed form value for this:  $$p = \left(\frac{G^{12}}{2\pi e^{\gamma}}\right)^{\pi^2/6},$$ where $G \approx 1.28243\ldots $ is the Glaisher constant, and $\gamma = \lim_{n \to \infty} H_n - \log n \approx 0.577216\ldots$ is Euler's gamma constant.
But if you just want to get bounds, I would try looking at improving the series convergence somehow, or using an integral approximation.
A: In short, take logs. Then you are left finding and bounding the series
$$ \sum_{n \geq 2} \frac{\log n}{n^2},$$
which is fairly rapidly convergent. The error from using the first $N$ terms will be approximately $1/N$, so using the first 20 terms will be enough to prove your inequalities, for instance.
