I'm trying to test
Sum[(-1)^(n + 1)/(3 n + 6 (-1)^n), {n, 0, Infinity}] == 1/3 (Log[2] - 1)
but I don't get an answer, just the reprint of the above. Am I doing something wrong?
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1 Answer
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Following Daniel Lichtblau's approach
t[n_] = (-1)^(n + 1)/(3 n + 6 (-1)^n);
t2[n_] = t[2 n] + t[2 n + 1] // Simplify[#, Element[n, Integers]] &
(* 1/(2*(-1 + n + 2*n^2)) *)
Sum[t2[n], {n, 0, Infinity}] // PowerExpand // Simplify
(* (1/3)*(-1 + Log[2]) *)
EDIT: Structuring as an equality test
Sum[t2[n], {n, 0, Infinity}] == (Log[2] - 1)/3
(* True *)
answered Oct 30, 2016 at 17:25
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$\begingroup$ Ah, very interesting, thank you. This is not a test but it is certainly useful to know how to do this manipulations when other forms of evaluation dont work directly. $\endgroup$ Oct 30, 2016 at 18:15
Infinity, notoo. If Mathematica returns aSumunchanged, it means that it cannot compute it symbolically. $\endgroup$Total@Table[(-1)^(n + 1)/(N[n] + 2 (-1)^n), {n, 0, 100000}]and checkN[Log[2]-1]. $\endgroup$Togetherto get a single term, then sum. This removes all the powers of -1.In[951]:= Simplify[ 1/3*Sum[Together[1/(n + 2) - 1/((n + 1) - 2)], {n, 0, Infinity, 2}]] Out[951]= 1/3 (1 - Log[2])(I tried with the correct sign but it does not simplify to the desired form). $\endgroup$