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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 !

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  • $\begingroup$ Do you mean Accumulate[Table[SeriesCoefficient[s, i], {i, 0, 29}]] ? $\endgroup$ – Mats Granvik Jan 5 '15 at 12:41
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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
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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 $-\pi / 2$.

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4
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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)*)
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