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I'm just curious if this would work. The first products I looked at are trivial. I just want to know if more can be said about this one:

$$f(x)=\prod_{n=1}^\infty\dfrac{1}{1+\large\frac{(-1)^n}{nx}}$$

Does this now converge?

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Please try it yourself. What do you thing SE is? A place where you can freely give other people questions that you have not thought about? Please dont. –  CBenni Jan 14 '13 at 16:35
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It's a valuable learning experience. I'm not afraid to ask stupid questions. –  user54358 Jan 14 '13 at 16:49
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Its not about bad questions. Please keep in mind other people take their own time to answer this. Maybe put some thought into it yourself. –  CBenni Jan 14 '13 at 17:16
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The most important thing is to never stop asking questions. I don't understand why you sir are telling me to do the exact opposite! –  user54358 Jan 14 '13 at 17:22
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I think what CBenni is saying is that you should tell us what you tried. Do you know any criterion for a product to converge? Have you tried to apply it? What difficulties did you encounter? Etc. –  jathd Jan 14 '13 at 18:13

1 Answer 1

So I just looked at convergence of the reciprocal.

$g(x)=\prod_{n=1}^\infty\left(1+\frac{(-1)^n}{nx}\right)$

Convergence of $g$ is equivalent to convergence of the sum

$\sum_{n=1}^\infty\frac{(-1)^n}{nx} = -\frac{1}{x}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n} = -\frac{ln2}{x}$

The last sum is the alternating harmonic series. Ok I don't know if this is correct. Because the sum is not defined for $x=0$ but the original product is undefined for $x=\frac{1}{n}$ where $n$ is odd.

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