Hi everyone,
I’m new with Mathematica and i don’t understand why this instruction doesn’t work:
Limit[Product[1 + Sqrt[k (n - k)]/n^2, {k, 1, n}], n -> \[Infinity]]
Can anyone help me please?
Hi everyone,
I’m new with Mathematica and i don’t understand why this instruction doesn’t work:
Limit[Product[1 + Sqrt[k (n - k)]/n^2, {k, 1, n}], n -> \[Infinity]]
Can anyone help me please?
Really a puzzle
k (n - k )/n^4 // Expand
-(k^2/n^4)+k/n^3
SetAttributes[f, {Listable, NumericFunction}]
f[n_] := With[{n1 = Floor[n]}, Product[1 + Sqrt[k/n1 (1 - k/n1)]/n1, {k, 1, n1}]]
In= Limit[f[n], n -> \[Infinity]]
Out = 1
With[{lim = Rationalize[f[10^6] // N, 1/100]},
Plot[ f[n], {n, 1, 50}, , PlotRange -> {1, 1.7},
ImageSize -> Small, Ticks -> {{0, 20, 40}, {0, lim }},
GridLines -> {None, {1, lim , 3/2}}]]
This is really a math question. Mathematica simply cannot do the limit.
First an approximation.
partials = Table[Product[1 + Sqrt[k (n - k)]/n^2, {k, 1, n}], {n, 20, 400, 20}];
N[partials]
(* Out[470]= {1.4681, 1.47549, 1.47761, 1.47858, 1.47913, 1.47948, \
1.47972, 1.4799, 1.48003, 1.48014, 1.48022, 1.48029, 1.48035, 1.4804, \
1.48044, 1.48048, 1.48051, 1.48054, 1.48057, 1.48059} *)
We can also get the limiting value numerically like so.
NLimit[Product[1 + Sqrt[k (n - k)]/n^2, {k, 1, n}], n -> Infinity]
(* Out[472]= 1.48097 *)
This happens to be a really good approximation to Exp[Pi/8]. There are a few steps required to show that that is in fact the limit.
Thanks to both of you.
I already proved that the limit was exp(pi/8) but I was curious about the capacities of mathematica to find it.
I didn’t know either the NLimit command which seems very useful.
Thank you very much for taking the time to answer me
Mathematica can certainly compute a relevant value.
Exp[Limit[Integrate[Sqrt[k (n - k)]/n^2, {k, 1, n},
Assumptions -> n > 1000], n -> Infinity]]
(* Out[478]= E^(\[Pi]/8) *)
Reducing the limit to that computation is a different matter.