Show that complex product does not converge

Dear **MHF** members,

I have the following product of complex numbers

$\displaystyle \prod_{j=1}^{\infty}\bigg[\frac{1}{2^{j}}+\mathrm{i}\bigg(1-\frac{1}{2^{j}}\bigg)\bigg]$

and Mathematica shows me that the product accumulates to 4 different points.

Each is located in different quadrant and they are outside of the circle with a radius of half.

However, I cannot compute them explicitly.

Please show me some directions.

Thank you.

**bkarpuz**

Re: Show that complex product does not converge

Hey bkarpuz.

With regards to your original thread title (i.e. convergence), you can show that if the sum of the logs converges then the original product converges.

See if you can prove convergence first because if you can't then you won't have a unique answer.

Re: Show that complex product does not converge

Quote:

Originally Posted by

**chiro** Hey bkarpuz.

With regards to your original thread title (i.e. convergence), you can show that if the sum of the logs converges then the original product converges.

See if you can prove convergence first because if you can't then you won't have a unique answer.

Thank you very much **chiro**.

I can simply prove that non-convergence (actually accumulation) as follows.

** Proof.** As we know for any complex number $\displaystyle z=x+\mathrm{i}y$,

$\displaystyle |x|+|y|\geq|z|\geq\max\{|x|,|y|\}$.

Let $\displaystyle x_{n}:=\frac{1}{2^{n}}$ and $\displaystyle z_{n}:=x_{n}+\mathrm{i}(1-x_{n})$,

then we have $\displaystyle |z_{j}|\geq|1-x_{j}|$, which yields

$\displaystyle \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|=\prod_{i=1}^ {\infty}|z_{i}|\geq\prod_{i=1}^{\infty}|1-x_{i}|\approx\ell:=0.2887880951$.

This shows that product cannot tend to $\displaystyle 0$ (i.e., $\displaystyle 0+\mathrm{i}0$).

On the other hand, we have

$\displaystyle \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|\leq\prod_{i= 1}^{\infty}(|x_{i}|+|1-x_{i}|)=1$

showing that the product can not also diverge (in the sense of complex modulus).

Then, the complex sequence $\displaystyle w_{n}:=\prod_{i=1}^{n}z_{i}$ lies in the annulus$\displaystyle \mathcal{A}:\ \ell\leq|z|\leq1$.

Due to Bolzano-Weierstrass theorem (considered separetely for the real and imaginary parts),

there exists a subsequence $\displaystyle w_{n_{k}}$ of $\displaystyle w_{n}$ which converges to some $\displaystyle w\in\mathcal{A}$.

Recall that $\displaystyle \lim_{n\to\infty}z_{n}=\mathrm{i}$ and $\displaystyle w_{n_{k}+p}=w_{n_{k}}\prod_{i=n_{k}+1}^{p}z_{i}$,

which shows that $\displaystyle \lim_{k\to\infty}w_{n_{k}+p}=\mathrm{i}^{p}w$ for $\displaystyle p=0,1,\cdots$.

Therefore, the product accumulates to at least (also at most) four different points $\displaystyle \pm w,\pm\mathrm{i}w$ since $\displaystyle w\neq0$. $\displaystyle \rule{0.2cm}{0.2cm}$

But I just wanted to know if it is possible to find (by hand) what $\displaystyle w$ is.

Add: Show that complex product does not converge

Quote:

Originally Posted by

**bkarpuz** ** Proof.** As we know for any complex number $\displaystyle z=x+\mathrm{i}y$,

$\displaystyle |x|+|y|\geq|z|\geq\max\{|x|,|y|\}$.

Let $\displaystyle x_{n}:=\frac{1}{2^{n}}$ and $\displaystyle z_{n}:=x_{n}+\mathrm{i}(1-x_{n})$,

then we have $\displaystyle |z_{j}|\geq|1-x_{j}|$, which yields

$\displaystyle \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|=\prod_{i=1}^ {\infty}|z_{i}|\geq\prod_{i=1}^{\infty}|1-x_{i}|\approx\ell:=0.2887880951$.

Addentum. Recall that

$\displaystyle \prod_{j=1}^{\infty}\bigg(1-\frac{\lambda^{2}}{(j\pi)^{2}}\bigg)=\frac{\sin( \lambda)}{\lambda}$.

(See Basel problem - Euler's Approach)

If $\displaystyle \lambda\geq\frac{3\pi}{2\sqrt{2}}$, we have $\displaystyle \lambda\geq\frac{j\pi}{\sqrt{2^{j}}}$ for all $\displaystyle j=1,2,\cdots$.

The sequence on the right-hand side increases for $\displaystyle j=1,2$,

reaches its maximum value $\displaystyle \frac{3\pi}{2\sqrt{2}}\approx3.33216$ at $\displaystyle j=3$,

and then decreasses for $\displaystyle j=4,5,\cdots$.

This yields

$\displaystyle 1-x_{j}=1-\frac{1}{2^{j}}\geq1-\frac{\lambda^{2}}{(j\pi)^{2}}$ for all $\displaystyle j=1,2,\cdots$

provided that $\displaystyle \lambda\geq\frac{3\pi}{2\sqrt{2}}$.

Letting $\displaystyle \lambda\geq\frac{3\pi}{2\sqrt{2}}$ and we have for the *real* product

$\displaystyle \prod_{j=1}^{\infty}(1-x_{j})=\prod_{j=1}^{\infty}\bigg(1-\frac{1}{2^{j}}\bigg)\geq\prod_{j=1}^{\infty}\bigg (1-\frac{\lambda^{2}}{(j\pi)^{2}}\bigg)=\frac{\sin( \lambda)}{\lambda}$,

which is strictly positive for $\displaystyle \lambda=2\pi+\frac{1}{2}\pi>\frac{3\pi}{2\sqrt{2}}$. $\displaystyle \rule{0.2cm}{0.2cm}$

Quote:

Originally Posted by

**bkarpuz** This shows that product cannot tend to $\displaystyle 0$ (i.e., $\displaystyle 0+\mathrm{i}0$).

On the other hand, we have

$\displaystyle \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|\leq\prod_{i= 1}^{\infty}(|x_{i}|+|1-x_{i}|)=1$

showing that the product can not also diverge (in the sense of complex modulus).

Then, the complex sequence $\displaystyle w_{n}:=\prod_{i=1}^{n}z_{i}$ lies in the annulus$\displaystyle \mathcal{A}:\ \ell\leq|z|\leq1$.

Due to Bolzano-Weierstrass theorem (considered separetely for the real and imaginary parts),

there exists a subsequence $\displaystyle w_{n_{k}}$ of $\displaystyle w_{n}$ which converges to some $\displaystyle w\in\mathcal{A}$.

Recall that $\displaystyle \lim_{n\to\infty}z_{n}=\mathrm{i}$ and $\displaystyle w_{n_{k}+p}=w_{n_{k}}\prod_{i=n_{k}+1}^{n_{k}+p}z_ {i}$,

which shows that $\displaystyle \lim_{k\to\infty}w_{n_{k}+p}=\mathrm{i}^{p}w$ for $\displaystyle p=0,1,\cdots$.

Therefore, the product accumulates to at least (also at most) four different points $\displaystyle \pm w,\pm\mathrm{i}w$ since $\displaystyle w\neq0$. $\displaystyle \rule{0.2cm}{0.2cm}$

But I just wanted to know if it is possible to find (by hand) what $\displaystyle w$ is.