By the Weierstrass factorizaion theorem we obtain that "f(z)=\\prod\\limits_{n=1}^{\\infty}E_n\\left(\\frac{z}{a_n}\\right)" is a entire function, such that "\\{z: f(z)=0\\}=\\{a_n\\}_{n\\in\\mathbb N}", and these zeros are simple, where "E_n(z)=(1-z)\\exp\\left(\\sum\\limits_{k=1}^n \\frac{z^k}{k}\\right)"
Indeed
1) "\\lim\\limits_{n\\to\\infty}|a_n|=+\\infty"
2)For every "r>0" there is "N" such that "\\left|\\frac{r}{a_n}\\right|<\\frac{1}{2}" for all "n>N", then"\\sum\\limits_{n=1}^{\\infty}\\left|\\frac{r}{a_n}\\right|^{n+1}=\\sum\\limits_{n=1}^N\\left|\\frac{r}{a_n}\\right|^{n+1}+\\sum\\limits_{n=N+1}^{\\infty}\\left|\\frac{r}{a_n}\\right|^{n+1}\\le"
"\\le\\sum\\limits_{n=1}^N\\left|\\frac{r}{a_n}\\right|^{n+1}+\\sum\\limits_{n=N+1}^{\\infty}\\left(\\frac{1}{2}\\right)^{n+1}", so "\\sum\\limits_{n=1}^{\\infty}\\left|\\frac{r}{a_n}\\right|^{p_n+1}" , where "p_n=n", is convergent series.
Then "f(z)=\\prod\\limits_{n=1}^{\\infty}E_{p_n}\\left(\\frac{z}{a_n}\\right)=\\prod\\limits_{n=1}^{\\infty}E_n\\left(\\frac{z}{a_n}\\right)" is a entire function with simple zeros "\\{a_n\\}_{n\\in\\mathbb N}".
Answer: "f(z)=\\prod\\limits_{n=1}^{\\infty}E_n\\left(\\frac{z}{a_n}\\right)=\\prod\\limits_{n=1}^{\\infty}E_n\\left(\\frac{z}{\\sqrt{n}}\\right)", where "E_n(z)=(1-z)\\exp\\left(\\sum\\limits_{k=1}^n \\frac{z^k}{k}\\right)"
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