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Amplifying on @Jinxed answer: \$Version "10.0 for Mac OS X x86 (64-bit) (December 4, 2014)" o[a_] := Sum[ x^(2 k + 1) Product[(2 i + 1)^2 - a^2, {i, 0, k - 1}]/((2 k + 1)!), {k, 0, Infinity}] o[a] Sin[a*ArcSin[x]]/a o[0] ArcSin[x] Limit[o[a], a -> 0] ArcSin[x] o[0] // Trace term[k_] = x^(2 k + 1) Product[(2 i + ...

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If you evaluate the term first, the result is Sin[a ArcSin[x]]/a which, when assigned to o[a_], will further evaluate to x for o[1]. So, the solution to your problem would be to Evaluate the term Then define o[a_] on the result.

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Expanding on the comment by @wxffles Sum[(-1)^k*Binomial[n, k]*Binomial[k, j], {k, 0, n}, GenerateConditions -> True] For the case when j == n, Sum[(-1)^k*Binomial[n, k]*Binomial[k, n], {k, 0, n}, GenerateConditions -> True] Combining these results sum[j_, n_] = Piecewise[{{Sum[(-1)^k*Binomial[n, k]*Binomial[k, j], {k, 0, n}], j ...

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You can use Piecewise to filter out any components you do not desire: (* your exclusion conditions *) conds={2-n/2<0, (* possibly more *) }; filter[c_]:=Piecewise[{{1,And@@c}},0] and then Sum[filter[conds] c[n]r^(2-n/2) + filter[conds] d[n]r^(-n/2),{n,Infinity}] This can easily be expanded for any conditions you might encounter.

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An straightforward way to put exclusions into sums is by using KroneckerDelta: excl = {4, 9, 14}; Sum[ Times @@ (1 - Thread[KroneckerDelta[n, excl]]) r^(1 - l[n]) DD[n], {n, 1, Infinity} ] $$\sum _{n=1}^{\infty } \text{DD}(n) (1-\delta _{4,n}) (1-\delta _{9,n}) (1-\delta _{14,n}) r^{1-l(n)}$$ To show that this also works when the summand ...

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Since you know your exclusions beforehand, you can proceed like this (for any number of exclusions!): excl={4,9,14}; (* just an example of more than one exclusion! *) Sum[r^(1-l[n]) DD[n],{n, Select[Range[1, Max@excl],!MemberQ[excl,#]&]}]+ Sum[r^(1-l[n]) DD[n],{n,Max@excl+1,Infinity}] This will filter out all your exclusions in one run. It will ...

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I think this is what you are after: mySum[s_, a_, b_]:= Sum[ 1./n p^(-n s), {n, b}, {p, Prime @ Range @ PrimePi @ a}] Edit: skipper inner Sum, thanks to kguler's comment.

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