Show that complex product does not converge

Dear **MHF** members,

I have the following product of complex numbers

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 ,

.

Let and ,

then we have , which yields

.

This shows that product cannot tend to (i.e., ).

On the other hand, we have

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

Then, the complex sequence lies in the annulus .

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

there exists a subsequence of which converges to some .

Recall that and ,

which shows that for .

Therefore, the product accumulates to at least (also at most) four different points since .

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

Add: Show that complex product does not converge

Quote:

Originally Posted by

**bkarpuz** ** Proof.** As we know for any complex number

,

.

Let

and

,

then we have

, which yields

.

Addentum. Recall that

.

(See Basel problem - Euler's Approach)

If , we have for all .

The sequence on the right-hand side increases for ,

reaches its maximum value at ,

and then decreasses for .

This yields

for all

provided that .

Letting and we have for the *real* product

,

which is strictly positive for .

Quote:

Originally Posted by

**bkarpuz** This shows that product cannot tend to

(i.e.,

).

On the other hand, we have

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

Then, the complex sequence

lies in the annulus

.

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

there exists a subsequence

of

which converges to some

.

Recall that

and

,

which shows that

for

.

Therefore, the product accumulates to at least (also at most) four different points

since

.

But I just wanted to know if it is possible to find (by hand) what

is.