Studying the convergence of an infinite product.

Question

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

Answer 1

Hint:

$$P_n = \left (\prod_{k=1}^n(1+(k/n)^2)\right)^{1/n}.$$

Answer 2

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 $$ \exp\left(-2+\frac{\pi}{2}+\log 2\right) = \color{red}{2e^{\frac{\pi}{2}-2}}\approx 1.3020546\ldots $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}