$\prod_{n=1}^{+\infty} \left(1-\frac{1}{2n^2}\right)>0$?

Question

How to prove that the infinite product $\prod_{n=1}^{+\infty} \left(1-\frac{1}{2n^2}\right)$ is positive ?

Thanks

Answer 1

Just prove the associated log series: $$\sum_{n=1}^{\infty}\log\Bigl(1-\frac1{2n^2}\Bigr)$$ converges. Observe the general term of this series $$\log\Bigl(1-\frac1{2n^2}\Bigr)\sim_\infty -\frac1{2n^2}.$$

Answer 2

$1-\dfrac{1}{n^2}<1-\dfrac{1}{2n^2}<\left(1-\dfrac{1}{n^2}\right)^2$,So you can see Its limit exist.

Sorry For I cannot use LaTex and My poor English.

Answer 3

Hint: Only the lower limit is a problem. Look e.g. at the product for $n\geq 2$, you could prove: $$ \prod_{n\geq 2} (1-\frac{1}{2 n^2}) \geq \prod_{n\geq 2} (1-\frac{1}{ n^2})=\prod_{n\geq 2} \frac{(n-1)(n+1)}{ n \cdot n}=\frac12$$ (the last being a telescopic product). So your product is $\geq \frac14$.

Answer 4

As you can see here $\sin z$ can be expressed by an infinite product, namely

$$\sin z=z \prod _{n=1}^{\infty } \left(1-\frac{z^2}{\pi ^2 n^2}\right)$$ Thus for $z=\dfrac{\pi }{\sqrt{2}}$ we get $$\sin\left(\dfrac{\pi }{\sqrt{2}}\right)=\dfrac{\pi }{\sqrt{2}}\,\prod _{n=1}^{\infty } \left(1-\frac{1}{2 n^2}\right)$$ hence $$\prod _{n=1}^{\infty } \left(1-\frac{1}{2 n^2}\right)=\dfrac{\sin\left(\dfrac{\pi }{\sqrt{2}}\right)}{\dfrac{\pi }{\sqrt{2}}}\approx 0.358$$

Hope this helps

Answer 5

The exact value of such product can be derived from the Weierstrass product for the sine function, as already shown by Raffaele. As an alternative approach, we may notice that $$ \prod_{n\geq 1}\left(1-\frac{1}{2n^2}\right)^2 = \frac{1}{4}\prod_{n\geq 2}\left(1-\frac{1}{n^2}+\frac{1}{4n^4}\right)\geq\frac{1}{4}\prod_{n\geq 2}\frac{n-1}{n}\cdot\frac{n+1}{n} $$ where the last product is a telescopic product: $$ \prod_{n=2}^{N}\frac{n-1}{n}\cdot\frac{n+1}{n}=\frac{N+1}{2N}\stackrel{N\to +\infty}{\longrightarrow}\frac{1}{2} $$ hence it follows that the value of the original product is $\color{red}{\large\geq\frac{1}{\sqrt{8}}}$.
Such lower bound turns out to be pretty accurate.

Answer 6

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

With $\ds{N \in \mathbb{N}_{\geq 1}}$:

\begin{align} \prod_{n = 1}^{N}\pars{1 - {1 \over 2n^{2}}} & = \prod_{n = 1}^{N}{2\pars{n - \root{2}/2}\pars{n + \root{2}/2} \over 2n^{2}} = {\pars{1 - \root{2}/2}^{\overline{N}}\pars{1 + \root{2}/2}^{\overline{N}} \over \pars{N!}^{2}} \\[5mm] & = {\Gamma\pars{N + 1 - \root{2}/2} \over \Gamma\pars{1 - \root{2}/2}N!}\, {\Gamma\pars{N + 1 + \root{2}/2} \over \Gamma\pars{1 + \root{2}/2}N!} \\[5mm] & = {1 \over \Gamma\pars{1 - \root{2}/2}\Gamma\pars{\root{2}/2}\root{2}/2}\, {\pars{N - \root{2}/2}! \over N!}\,{\pars{N + \root{2}/2}! \over N!} \\[5mm] & = {\root{2} \over \pi/\sin\pars{\pi\root{2}/2}}\, {\pars{N - \root{2}/2}! \over N!}\,{\pars{N + \root{2}/2}! \over N!} \end{align}

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Note that \begin{align} {\pars{N + \alpha}! \over N!} & \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, {\root{2\pi}\pars{N + \alpha}^{N + \alpha + 1/2}\expo{-\pars{N + \alpha}} \over {\root{2\pi}N^{N + 1/2}\expo{-N}}} = {N^{N + \alpha + 1/2}\pars{1 + \alpha/N}^{N + \alpha + 1/2}\expo{-\alpha} \over N^{N + 1/2}} \\[5mm] & \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, N^{\alpha} \end{align}

such that

\begin{align} \prod_{n = 1}^{\infty}\pars{1 - {1 \over 2n^{2}}} & = \bbx{{\root{2}\sin\pars{\sqrt{2}\,\pi/2} \over \pi}} \approx 0.3582 \end{align}