Studying the convergence of an infinite product. I have the following sequence,
\begin{align*}
 P_n=\displaystyle \cfrac{1}{n^2}{\prod_{k=1}^{n}(n^2+k^2)^\frac{1}{n}} \:\:\:\: \:\:   n\geq 1 
\end{align*}
The sequence seems to converge toward zero. But I have a hard time proving it. My strategy is to use a the following theorem. 
Theorem. If $a_n ≥ 0 $ for all $n ≥ 1$, then the infinite product $\displaystyle\prod_{n=1}^{\infty}
(1 + a_n)$ converges if and only if the infinite
series $\displaystyle\sum_{n=1}^{\infty}
a_n$ converges.
I am trying to decompose the product in partial sums but the $k$ is giving me trouble. Any thoughts would be appreciated
 A: Hint:
$$P_n = \left (\prod_{k=1}^n(1+(k/n)^2)\right)^{1/n}.$$
A: Further hint:
$$ \log P_n = \frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k^2}{n^2}\right)\stackrel{\text{Riemann sums}}{\longrightarrow}\int_{0}^{1}\log(1+x^2)\,dx =-2+\frac{\pi}{2}+\log 2$$
gives that $P_n$ does not converge toward zero, but toward
$$ \exp\left(-2+\frac{\pi}{2}+\log 2\right) = \color{red}{2e^{\frac{\pi}{2}-2}}\approx 1.3020546\ldots $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[15px,#ffc]{\left.{1 \over n^{2}}\prod_{k = 1}^{n}\pars{n^{2} + k^{2}}^{1/n}
\right\vert_{\ n\ \geq\ 1}} =
{1 \over n^{2}}\braces{\bracks{\prod_{k = 1}^{n}\pars{k + \ic n}}
\bracks{\prod_{q = 1}^{n}\pars{q - \ic n}}}^{1/n}
\\[5mm] = &\
{1 \over n^{2}}\verts{\prod_{k = 1}^{n}\pars{k + \ic n}}^{\ 2/n} =
{1 \over n^{2}}\verts{\pars{1 + \ic n}^{\large\overline{n}}}^{\ 2/n} =
{1 \over n^{2}}\verts{\Gamma\pars{\bracks{1 + \ic n} + n} \over
\Gamma\pars{1 + \ic n}}^{\ 2/n}
\\[5mm] = &\
{1 \over n^{2}}\verts{\pars{\ic n + n}! \over \pars{\ic n}!}^{\ 2/n}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{1 \over n^{2}}\verts{\root{2\pi}\pars{\ic n + n}^{\ic n + n + 1/2}
\expo{-\ic n - n} \over \root{2\pi}\pars{\ic n}^{\ic n + 1/2}
\expo{-\ic n}}^{\ 2/n}
\\[5mm] = &\
{1 \over n^{2}}\verts{{n^{\ic n + n + 1/2}\pars{1 + \ic}^{\ic n + n + 1/2}
 \over
n^{\ic n + 1/2}\,\ic^{\ic n + 1/2}}\,\expo{-n}}^{\ 2/n} =
{1 \over n^{2}}\verts{{n\pars{1 + \ic}^{\ic + 1 + 1/\pars{2n}}
 \over
\ic^{\ic + 1/\pars{2n}}}\,{1 \over \expo{}}}^{\ 2}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, &\
\verts{\pars{1 + \ic}^{1 + \ic} \over \ic^{\ic}}^{2}\,{1 \over \expo{2}} =
\verts{2^{1/2 + \ic/2}\expo{\ic\pi/4}\expo{-\pi/4} \over
\expo{-\pi/2}}^{2}\,{1 \over \expo{2}} =
\verts{2^{1/2}\expo{\pi/4} }^{2}\,{1 \over \expo{2}}
\\[5mm] = &\
\bbox[#ffc,15px,border:1px groove navy]{2\expo{\pi/2 - 2}}\
\approx\ 1.3021
\end{align}
