I would like to know an simple way to justify the convergence of the infinite product

$$\xi:=\prod_{n=2}^\infty\left(1-\frac{1}{n^2}\right)^{i^n},\tag{1}$$ where $i=\sqrt{-1}$ is the imaginary unit.

My question arises when I wondered about a similar question than [1].

Question. Please provide a justification that $\xi$ converges*, and the following inequality between its real part and its imaginary part $$\Re \xi>\Im \xi\tag{2}$$ holds. Many thanks.

*Only is required to show that the infinite product is well-defined, that is convergent (isn't required a closed-form for $\xi$, if my first question is in the literature you can to assume that it converges and solve the second question). My second question arises from a calculation that I did with a CAS.

I think thus that we need to study the partial products of our infinite product and use the perioricity of the sequence $i^n$, as $n\geq 1$ runs over positive integers.


[1] Problem 3673, Crux Mathematicorum, Vol. 38(8), October 2012.


Rewrite your product using $z^w=e^{w \log z}$ (for positive real $z$). $$ \prod_{n=2}^{\infty} \left( 1- \frac{1}{n^2} \right)^{i^n} = \prod_{n=2}^{\infty} e^{\log \left( 1- \frac{1}{n^2} \right) i^n} = e^{\sum_{n=2}^{\infty} \log \left( 1- \frac{1}{n^2} \right) i^n}$$ This series is absolutely convergent, thus it is convergent. Indeed $$\sum_{n=2}^{\infty} \left| \log \left( 1- \frac{1}{n^2} \right) i^n \right| = \sum_{n=2}^{\infty}\left| \log \left( 1- \frac{1}{n^2} \right) \right| < \infty$$ is convergent since $\log \left( 1- \frac{1}{n^2} \right) \sim -\frac{1}{n^2}$.

This proves that the original product is absolutely convergent. Thus, you can rearrange the factors arbitrarily without affecting the value of the product. Moreover, every "subproduct" will be convergent and well-defined.

In particular, you can separate even terms and odd terms. Recalling that $i^{2n}= (-1)^n$ and $i^{2n+1}= (-1)^ni$ $$\prod_{n=2}^{\infty} \left( 1- \frac{1}{n^2} \right)^{i^n} = \prod_{n=1}^{\infty}\left( 1- \frac{1}{(2n)^2} \right)^{(-1)^n} \cdot \prod_{n=1}^{\infty}\left( 1- \frac{1}{(2n+1)^2} \right)^{(-1)^ni}$$ let's call $K=\prod_{n=1}^{\infty}\left( 1- \frac{1}{(2n)^2} \right)^{(-1)^n}>0$, which is not relevant for the answer of the second question. So you only have to consider the second product. Again, $$\prod_{n=1}^{\infty}\left( 1- \frac{1}{(2n+1)^2} \right)^{(-1)^ni} = \exp \left( \sum_{n=1}^{\infty} (-1)^n \log \left( 1- \frac{1}{(2n+1)^2} \right) i \right) = \cos H + i \sin H$$ where $$H=\sum_{n=1}^{\infty} (-1)^n \log \left( 1- \frac{1}{(2n+1)^2} \right)$$

So you are left to prove that $$\cos H > \sin H$$ This follows by the fact that $0 < H <\pi/4$. Indeed, the series defining $H$ is an alternating series, giving you the estimate of $H$ bounded between the first two partial summations: $$\log \left( 1- \frac{1}{5^2} \right) -\log \left( 1- \frac{1}{3^2} \right) <H< -\log \left( 1- \frac{1}{3^2} \right) $$ i.e. $$\log(27/25) < H < \log(9/8)$$ And this concludes the proof since $0 < \log (27/25)<\log (9/8) < \pi/4$.

  • $\begingroup$ Many thanks, it is sad that didn't remember that the argument of your first paragraph works for our complex numbers ( otherwise I would have included it). $\endgroup$
    – user243301
    Jun 25 '18 at 8:48
  • $\begingroup$ I've shortened your answer to the second question. It's too long for a comment, so I've included it as an answer, $\endgroup$
    – zhw.
    Jun 25 '18 at 23:14

This is too long for a comment. I just want to point out we can shorten @crostul's answer for the second question. Rearrangements, subproducts: we don't need to worry about them. Just define

$$S = \sum_{n=2}^{\infty}i^n\log (1-1/n^2)=A+iB.$$

Our infinite product equals $e^{A+iB} = e^{A}e^{iB}.$ So if suffices to show $0<B<\pi/4.$ But with a little work you see

$$B= \ln(9/8) - \ln(25/24) + \ln(49/48) - \cdots$$

In absolute value these terms are strictly decreasing and $\to 0.$ Thus, as is well known, $B$ lies between $0$ and $\ln (9/8).$ But since $\ln (1+u)<u$ for $u>0,$ we have

$$\ln (9/8) =\ln (1+1/8) < 1/8 < \pi/4,$$

and we're done.

  • $\begingroup$ Many thank you very much and to the other user answering the question, really, for sharing these calculations. $\endgroup$
    – user243301
    Jun 26 '18 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy