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Convert to polynom converts a Taylor series to a polynomial. If series is not a Taylor series then the conversion to sum-of-products form takes place but the result is not of type polynom.

$s≔\mathrm{series}\left(\sin \left(x\right)\,x\,5\right)$

$s≔x-\frac{1}{6}{x}^3+O\left(x^5\right)$

$\mathrm{type}\left(s\,\mathrm{polynom}\right)$

$\mathrm{false}$

$p≔\mathrm{convert}\left(s\,\mathrm{polynom}\right)$

$p≔x-\frac{1}{6}{x}^3$

$\mathrm{type}\left(p\,\mathrm{polynom}\left(\mathrm{anything}\,x\right)\right)$

$\mathrm{true}$

$t≔\mathrm{series}\left(\frac{\exp \left(x\right)}{x}\,x\,3\right)$

$t≔x^{\mathrm{-1}}+1+\frac{1}{2}x+O\left(x^2\right)$

$q≔\mathrm{convert}\left(t\,\mathrm{polynom}\right)$

$q≔\frac{1}{x}+1+\frac{x}{2}$

$\mathrm{type}\left(q\,\mathrm{polynom}\left(\mathrm{anything}\,x\right)\right)$

$\mathrm{false}$

$\mathrm{type}\left(q\,\mathrm{`+`}\right)$

$\mathrm{true}$