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I'm studying this series:

enter image description here

I know that there is this relation:

enter image description here

enter image description here

($\ell$ is the common limit of the two series)

Do you know a Mathematica function that could prove this relation?

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  • $\begingroup$ You should look at RSolve , although on first pass it doesn't seem to be able to solve this. $\endgroup$
    – george2079
    Commented Feb 20, 2018 at 19:43
  • $\begingroup$ I think you need a0,b0 to have the same sign BTW. $\endgroup$
    – george2079
    Commented Feb 20, 2018 at 19:49
  • $\begingroup$ Yes a0 and b0 are > 0. But, did you manage to have the result? $\endgroup$
    – nolwww
    Commented Feb 20, 2018 at 20:53
  • $\begingroup$ @george2079 I had a new picture with a new relation that might help. $\endgroup$
    – nolwww
    Commented Feb 20, 2018 at 20:56
  • $\begingroup$ I confirmed your result numerically. That's not a proof though. re: edit you should see what RSolve does with that. $\endgroup$
    – george2079
    Commented Feb 20, 2018 at 22:39

1 Answer 1

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This is not a proof but rather an empirical test

Clear[a, b]

#[{a, b}] & /@ {Mean, HarmonicMean, GeometricMean}

(* {(a + b)/2, 2/(1/a + 1/b), Sqrt[a b]} *)

Since the terms converge, then you can used FixedPoint to find the limit

gm[a_?Positive, b_?Positive] :=
 FixedPoint[{Mean[#], HarmonicMean[#]} &, {a, b}]

Testing whether the limits are the same and equal to the GeometricMean for 10,000 pairs of random reals

And @@ Table[
  {a, b} = RandomReal[{10^-9, 100}, 2, WorkingPrecision -> 15];
  g = gm[a, b];
  g[[1]] == g[[2]] == GeometricMean[{a, b}],
  {10000}]

(* True *)

You can use FixedPointList to look at the convergence step-by-step

FixedPointList[{Mean[#], HarmonicMean[#]} &, {5., 79.}]

(* {{5., 79.}, {42., 9.40476}, {25.7024, 15.3682}, {20.5353, 19.2352}, 
    {19.8852, 19.864}, {19.8746, 19.8746}, {19.8746, 19.8746}, 
    {19.8746, 19.8746}, {19.8746, 19.8746}} *)

GeometricMean[{5., 79.}]

(* 19.8746 *)
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2
  • $\begingroup$ I’m sorry but I don’t understand how it demonstrate the equality that I have to show ? $\endgroup$
    – nolwww
    Commented Feb 21, 2018 at 23:27
  • $\begingroup$ This shows that the common limit of the two series is the GeometricMean, i.e., l = Sqrt[a0*b0] $\endgroup$
    – Bob Hanlon
    Commented Feb 22, 2018 at 1:07

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