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This could be something which is already somewhere in the website, but I am unable to locate any.

Prove $$\prod_{n=1}^{\infty} (1-z^n)$$ converges absolutely and uniformly on each compact subset of $\biggr[{|z|<1}\biggr]$.

What about $\biggr[{|z|>1}\biggr]$?

I have a feeling this need to be done with using some logarithm expansion and using Weierstrass theorem. But I do like to see more ideas of proof. I am not sure on using those ideas either.

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Try taking $z=1+\epsilon$. What happens to the size of the terms in the product? – Eric Naslund Dec 23 '12 at 5:03
For more references see 'Euler function'. – Raymond Manzoni Dec 23 '12 at 9:10

You should use the definition of absolute convergence of a product: $\prod(1+y_n)$ converges absolutely if and only if $\sum|y_n|$ converges. You can also use the fact that a necessary condition for a product $\prod(1+y_n)$ to converge is that $y_n\to0$. These two statements should allow you to prove the first statement and show that in the second case, the product diverges.

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So the hint given by @Eric Naslund make sense now. Thanks Greg Martin. – Deepak Dec 23 '12 at 6:59
You're welcome! This is a good example to show why the fundamentals of a subject are so important. We can't treat every problem like a brand new mystery unto itself; we need to build up basic techniques to run new problems by, to sort the easily solved ones from the ones that require deeper though. – Greg Martin Dec 23 '12 at 14:50

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