I'm just curious because I was trying to come up with a weird function with weird discontinuities. Then I thought
$$f(x)=\prod_{n=1}^\infty \dfrac{1}{1-\frac{1}{nx}}$$
So what's this product?
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Try taking log(f(x)) and see what happens – Geoff Robinson Jan 14 '13 at 10:29
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1$\frac{n}{nx-1} = \frac{1}{x-1/n}$ so your product is zero except when $x = 1/k$ for some integer $k$ - then your product is infinite. – John Dvorak Jan 14 '13 at 10:30
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@Jan: I'd say for $x = 1/k$ it's undefined and that answer also depends on whether $x > 1$ or not. Still, the answer is trivial in all cases. – Marek Jan 14 '13 at 10:37
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Ok thanks it makes my function boring then. What if I have $\Pi_{n=1}^\infty\frac{nx}{nx-1}$ instead. Can I say more about this one? – user56766 Jan 14 '13 at 10:39
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@user56766 rewrite that to $\Pi^\inf_n=1\frac{1}{1-1/nx}$; this one looks more interesting – John Dvorak Jan 14 '13 at 10:40
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2Please ask a new question. – Did Jan 14 '13 at 10:44
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This was an answer to the OP's first question which has since been edited, which asked about the product $\displaystyle \prod_{n=1}^{\infty} \dfrac{n}{nx-1}$.
Consider the possible values of $x$ in cases.
If $|x|>1$, then for sufficiently large $n$, you'll have $|nx-1|>a|n|$ for some $a$ with $|a|>1$, so for these values of $x$, $f(x) = 0$.
For $-1 \le x < 0$, we have $$\dfrac{n}{nx-1} = -\dfrac{n}{n|x|+1} \to -\dfrac{1}{|x|} \le -1$$ and so the product doesn't converge.
Writing $\dfrac{n}{nx-1} = \dfrac{1}{x-\frac{1}{n}}$ makes it clear that for any $0 \le x < 1$ the function diverges, since you're taking the product of an infinite sequence of numbers $>1$ which does not converge to $1$.
Finally, $f(1)$ is undefined since one of the terms of your product is $\dfrac{1}{x-1}$.
In summary: $f(x)=0$ for $|x|>1$ and is undefined for $|x|<1$.
Clive Newstead
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For amended question. When $x >1,$ the log of the partial product to $k$ terms is $\sum_{n=1}^{k} \log(1 - \frac{1}{nx}),$ Let $r$ be the smallest integer with $rx >1.$ Then the partial product is greater than $\sum_{n=r}^{k} \frac{1}{nx}$ which tends to infinity as $k$ increases, since the harmonic series diverges. Hence the product diverges for $x > 1.$ If $x = \frac{1}{n}$ for any positive integer $n,$ then the product is not defined for that $x.$ For other values of $x <1,$ the product diverges by an argument similar to that at the beginning.
Geoff Robinson
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