I made the computation
ClearAll["Global`*"];
r = Sum[1/2^(k*n/(k + n)), {k, 1, 2*n}, Assumptions -> n ∈ Integers && n > 0]
and got
(1-4^(-(n^2/(n+Sum`SumqBaseDump`u$274844))))/(-1+2^(n/(n+Sum`SumqBaseDump`u$274844)))
I can find nothing about SumqBaseDump in the Mathematica documentation.
Addition. It should be noticed DiscreteLimit works with r:
DiscreteLimit[r, n -> Infinity]
1
Sumto me. – m_goldberg Oct 31 '18 at 12:13Assumptions -> n ∈ Integers && n > 0isn't necessary to produce the result. – xzczd Oct 31 '18 at 12:14RSolve[{a[m + 1] == a[m] + (1/2)^(n m /(n + m)), a[1] == 1/Sqrt[2]}, a[m], m]. – b.gates.you.know.what Oct 31 '18 at 14:11Sum[1/2^(1/(1/n + 1/k)), {k, 1, 2*n}]avoids this specific bug. – Wouter Jan 08 '19 at 17:26