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I need following

Series[Exp[-Sqrt[(a/p)]] HypergeometricU[n, m,b Sqrt[a/p]], {p,\[Infinity], 1}]

where n and m are any positive integers which do not have any relationship.

When I run this on Mathematica, it gives answer but only for non-integer n and m because it includes Gamma[1 - m + n] which has ComplexInfinity for negative 1 - m + n.

How I can run above restricting answer to be positive integers n and m?

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    $\begingroup$ Did you try the Assumptions option for Series ? $\endgroup$ – Lotus Mar 2 '17 at 5:07
  • $\begingroup$ @Lotus: It doesn't work. $\endgroup$ – user64494 Mar 2 '17 at 9:38
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The following works in Mathematica 11.

Series[(Exp[-Sqrt[(a/p)]] HypergeometricU[n, m, b Sqrt[a/p]]) /. {m ->     2, n -> 3},
{p, \[Infinity], 1}]

$$\frac{\sqrt{p}}{2 \sqrt{a} b}+\left(\log \left(\sqrt{a} b\right)-\frac{1}{2 b}+\frac{1}{2} \log \left(\frac{1}{p}\right)+\gamma +\frac{1}{2}\right)+\sqrt{\frac{1}{p}} \left(-\sqrt{a} \left(\log \left(\sqrt{a} b\right)+\frac{1}{2} \log \left(\frac{1}{p}\right)+\gamma +\frac{1}{2}\right)+\frac{1}{4} \left(3 \sqrt{a} b \log \left(\frac{1}{p}\right)-4 \sqrt{a} b+6 \gamma \sqrt{a} b+6 \sqrt{a} b \log \left(\sqrt{a} b\right)\right)+\frac{\sqrt{a}}{4 b}\right)+\frac{6 a b^3 \log \left(\frac{1}{p}\right)-15 a b^3+12 \gamma a b^3+12 a b^3 \log \left(\sqrt{a} b\right)-9 a b^2 \log \left(\frac{1}{p}\right)+12 a b^2-18 \gamma a b^2-18 a b^2 \log \left(\sqrt{a} b\right)+3 a b \log \left(\frac{1}{p}\right)+3 a b+6 \gamma a b+6 a b \log \left(\sqrt{a} b\right)-a}{12 b p}+O\left(\left(\frac{1}{p}\right)^{3/2}\right) $$

Maple confirms it.

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  • $\begingroup$ Can't we keep m and n as parameters? Not just specific values. $\endgroup$ – Frey Mar 2 '17 at 8:50
  • $\begingroup$ @Frey: As you wrote in your question, we can't use the output of Series[Exp[-Sqrt[(a/p)]] HypergeometricU[n, m,b Sqrt[a/p]], {p,[Infinity], 1}]. Neither taking of limits nor assumptions work. $\endgroup$ – user64494 Mar 2 '17 at 9:35

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