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Construct a function with zero at $z=0$ and zeros at $z=-n$ with multiplicities $n$.

My answer is $$f(z) = z\prod_{n=1}^{\infty}\left[E_n\left(-\frac zn\right)\right]^n,$$ where $E_n(z)=(1-z)\exp\left(\sum_{k=1}^n\frac{z^k}k\right)$.

Is that right? And does the product converge?

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    $$E_n(z)$$ is $$E_n(z)=(1-z)e^{z+z^2/2+z^3/3+...+z^n/n}$$2012-03-16
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    Why not something like $f(z)=z\prod_{n=1}^\infty a_n \left(1+\frac{z}n\right)^n$ for some suitable choice of constants $a_n$ (so that we get nice convergence)? As long as $\lim_{n\rightarrow\infty}a_n<\frac1e$ we should be okay.2012-03-16
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    @bgins: OP's is the form in the [Weierstrass factorization](http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem), which ensures that the terms in the product tend to $1$ in the limit (notice the polynomial in the exponential is a truncation of the Maclaurin expansion of $\log(1-z)^{-1}$). The form you cite won't work because $(1+\frac{z}{n})^n\to e^z$, so the product won't converge unless $z$ is a nonpositive integer. | *Edit*: Even with $a_n$ coefficients it won't work because $e^z$ varies with $z$.2012-03-16

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