$\sum _{k=0}^{\infty } -\frac{2^{-2 k} j^{2 k} \Gamma \left(\frac{1}{2}-k,10\right)}{\Gamma (1+k)}+\sum _{k=0}^{\infty } -\frac{2 j^{2 k} \Gamma (-2 k)
\sin (k \pi )}{\sqrt{\pi }}$
Sum[#, {k, 0, Infinity}] & /@ (Gamma[-k + 1/2, 0, 10]/k!*(j/2)^(2 k) //
FunctionExpand)
(*Sum[-((j^(2*k)*Gamma[1/2 - k, 10])/(2^(2*k)*Gamma[1 + k])), {k, 0, Infinity}] +
Sum[(-2*j^(2*k)*Gamma[-2*k]*Sin[k*Pi])/Sqrt[Pi], {k, 0, Infinity}]*)
First series we can compute from identity
$\Gamma (a,z)=\int_z^{\infty } t^{a-1} \exp (-t) \, dt$
Integrate[
Sum[-((2^(-2 k) j^(2 k) (t^(a - 1) Exp[-t] /. a -> 1/2 - k))/
Gamma[1 + k]), {k, 0, Infinity}], {t, 10, Infinity}, Assumptions -> j >= 0]
(*-(1/2) E^(-I j) Sqrt[\[Pi]] (Erfc[(20 - I j)/(2 Sqrt[10])] +
E^(2 I j) Erfc[(20 + I j)/(2 Sqrt[10])])*)
Second series we can compute like this:
L = Table[
Limit[-((2 j^(2 k) Gamma[-2 k] Sin[k \[Pi]])/Sqrt[\[Pi]]), k -> m,
Assumptions -> j >= 0], {m, 0, 10}]
Sum[FindSequenceFunction[L, k] // FunctionExpand, {k, 0, Infinity}]
(*Sqrt[\[Pi]] Cos[j]*)
Regards M.I.
