I am computing the Series expansion (Lauren series) of an integral and I want to sum up the coefficients of the series.
a = 1; s = Series[ Integrate[2*x*ArcSin[a/x] - 2*a*Sqrt[1 - (a/x)^2], x], {x, Infinity,29}] Table[SeriesCoefficient[s, i], {i, 0, 29}] I am able to generate the Coefficient list but How to sum up them? Could someone help me please? Thanks a lot !
Simply replace Table by Sum:
a = 1; s = Series[ Integrate[2*x*ArcSin[a/x] - 2*a*Sqrt[1 - (a/x)^2], x], {x, Infinity,29}] Sum[SeriesCoefficient[s, i], {i, 0, 29}] Or if you want to keep the table, use Total:
a = 1; s = Series[ Integrate[2*x*ArcSin[a/x] - 2*a*Sqrt[1 - (a/x)^2], x], {x, Infinity,29}] t = Table[SeriesCoefficient[s, i], {i, 0, 29}] Total @ t If you are interested in the analytic form for the series coefficient $c[n]$ of $x^{-n}$, it is (I think) $c[n]=-\frac{4~\Gamma[n/2]}{n~(n+2)~\sqrt{\pi}~\Gamma[(n+1)/2]}$, for $n=1,3,5,7,...$. (Gory details available on request).
Finding the sum of coefficients from $n=1$ to arbitrary odd upper limit $m$ has an analytic form in terms of $m$ returned by the following Sum.
FullSimplify[Sum[c[n],{n,1,m,2}],Assumptions -> m>0] The infinite sum of all coefficients is given by
Sum[c[n],{n,1,Infinity,2}] which evaluates to $-\pi / 2$.
It looks like you're trying to approximate the result using a partial sum of the first 30 Laurent coefficients. But you can actually get the infinite sum with far less code by substituting $x=1$:
Limit[Integrate[2*x*ArcSin[a/x] - 2*a*Sqrt[1 - (a/x)^2], x], x -> 1] (*-(π/2)*) 