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

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

Thanks

• Since each $1-\frac{1}{2n^2}$ is $>0$? – awllower Aug 25 '17 at 9:10
• @awllower it's an infinite product. – M. Houng Aug 25 '17 at 9:11
• So it remains to prove the limit exists, and I think you should specify this. – awllower Aug 25 '17 at 9:11
• Even if the limit exists, it could be 0. – M. Houng Aug 25 '17 at 9:13
• Total lack of context. For example, do you know any method to prove that any infinite product is positive? – Did Aug 25 '17 at 9:27

## 6 Answers

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 $\geq \frac14$.

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 $$\log\Bigl(1-\frac1{2n^2}\Bigr)\sim_\infty -\frac1{2n^2}.$$

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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}

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}

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

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.

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

Sorry For I cannot use LaTex and My poor English.