11
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @Kuba: Thank you for your edit. $\endgroup$ – user64494 Oct 31 '18 at 11:54
  • 9
    $\begingroup$ Looks like a bug in Sum to me. $\endgroup$ – m_goldberg Oct 31 '18 at 12:13
  • 1
    $\begingroup$ Same behavior in v8.0.4 and v9.0.1. Assumptions -> n ∈ Integers && n > 0 isn't necessary to produce the result. $\endgroup$ – xzczd Oct 31 '18 at 12:14
  • $\begingroup$ @Mariyusz Iwaniuk: And no constructive result. $\endgroup$ – user64494 Oct 31 '18 at 12:14
  • 1
    $\begingroup$ Same with RSolve[{a[m + 1] == a[m] + (1/2)^(n m /(n + m)), a[1] == 1/Sqrt[2]}, a[m], m]. $\endgroup$ – b.gates.you.know.what Oct 31 '18 at 14:11

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