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3

Sum[(x^n)/(n + a), {n, 0, Infinity}, GenerateConditions -> True] (* ConditionalExpression[HurwitzLerchPhi[x, 1, a], Abs[x] <= 1 && x != 1] *)


2

You can do this; if you want even without touching anything manually but you need to transform some expressions. First, let's make a rule from your substitution: rule = y -> -r^2 + 2*r*Cos[\[Theta]] It reads as "y is replaced by ...". For the first substitution, this wouldn't help us because we need the other way around. Additionally, you have to ...


1

Some points: You are integrating a function of InterpolatingFunction. See this thread for guidance how to achieve maximum precision in this situation using NIntegrate. You are using Sum for summing up imprecise numbers which is the worst way to do this as demonstrated here. Use Total with option "CompensatedSummation" -> True instead. Avoid using ...


5

You already have the answer, your mistake was to expand around the point the function diverges, i.e Log[0]. Try changing the second element of the second argument, for example: FullSimplify@PadeApproximant[Log[x], {x, 1, {6, 6}}] (7 (-1 + x) (1 + x) (7 + x (132 + x (382 + x (132 + 7 x)))))/(10 (1 + x (36 + x (225 + x (400 + x (225 + x (36 + x))))))) ...



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