# Convergence of $\left\{n!\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}\right\}$ as $n\to \infty$ using Mathematica or otherwise

Using Mathematica or otherwise, I need to find the convergence or divergence of$$\lim_{n\to\infty}\left\{n!\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}\right\}$$ where $$\{x\}$$ denotes the fractional part of $$x$$ and $$n!$$ is the factorial of $$n$$.

We have $$\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}=H^{(\frac{3}{2})}_{n!}$$ where $$H^{(r)}_{n}$$ is the generalized harmonic number.
So we have $$\lim_{n\to\infty}\left\{n!\sum_{k=1}^{n!}\frac{1}{k^{\frac{3}{2}}}\right\}=\lim_{n\to\infty}\left\{n!H^{(\frac{3}{2})}_{n!}\right\}$$ I have observed the behavior of the fractional part using Wolfram Alpha. The plot shows that $$\left\{n!H^{(\frac{3}{2})}_{n!}\right\}$$ oscillates between $$y=0$$ and $$y=1$$. So my gut feeling is that the sequence $$\left\{n!H^{(\frac{3}{2})}_{n!}\right\}$$ is not convergent. I tried the following code on Wolfram Cloud as well:

 a // ClearAll;
a[n_] /; MemberQ[Stack[], Limit] := FractionalPart[n!*HarmonicNumber[n!,5]];
SumConvergence[a[n], n]


But I am getting output as:

 SumConvergence[a[n],n]


Again I tried

 SumConvergence[FractionalPart[n!*HarmonicNumber[n!,5]], k, Method->"IntegralTest"]


But I am getting the output as

 SumConvergence[FractionalPart[n! HarmonicNumber[n!,5]],k,Method->IntegralTest]


Please give a code for the convergence or divergence of the sequence. Any help would be highly appreciated.

• I'd recommend asking this question on the math SE because it does not seem suited for Mathematica analysis. Commented 2 days ago

Let's change n! to nfact.

sum = Sum[1/k^(3/2), {k, nfac}]
(* Out[66]= HarmonicNumber[nfac, 3/2] *)

InputForm[ser = Series[nfac*sum, {nfac, Infinity, 1}]]
(* Out[70]//InputForm=
SeriesData[nfac, Infinity, {Zeta[3/2], -2, 0, 1/2}, -2, 3, 2] *)


So the product is Zeta[3/2]*n! -2*Sqrt[n!]+... where the elided part is all terms going to zero. Notice that Sqrt[n!] is an algebraic number for all integer n. If zeta(3/2) is transcendental (unknown) then there can be no limit to the fractional part (it will give, as @roman noted, a sequence uniformly distributed mod 1). If it's algebraic then I've no idea what can be shown, other than it "probably" still won't have a limit.

• $(+1)$. Thanks. Please let me know if anything more comes up.
– Max
Commented yesterday
• If $\zeta(3/2)$ is algebraic, I'll start a podcast about paint drying. Commented 54 mins ago

Let's define some functions to evaluate these numbers at high accuracy:

$MaxExtraPrecision = ∞; f[n_Integer?Positive] := FractionalPart[n! HarmonicNumber[n!, 3/2]] g[n_Integer?Positive] := g[n] = N[f[n], 100] // N  Show the sequence: DiscretePlot[g[n], {n, 1000}]  The numbers seem to be pretty random and spread uniformly over $$[0,1]$$: Histogram[Array[g, 1000], {0, 1, 1/10}]  A Kolmogorov–Smirnov test supports this observation: KolmogorovSmirnovTest[Array[g, 1000], UniformDistribution[]] (* 0.24534 *)  If it is true that the sequence $$f_n$$ is quasi-random and uniform on $$[0,1]$$, then it won't converge. • Thanks. ($+1$). How to show that the sequence$f_n$is quasi-random and uniform on$[0,1]\$?
– Max
Commented 2 days ago
• No idea – that's why I suggested going to the math stackexchange. The Kolmogorov–Smirnov test is merely an indication that it looks quasi-random. Commented 2 days ago
• What do you mean by quasi- random and uniform
– Max
Commented 2 days ago
• I mean a low-discrepancy sequence. Quasi-random, for practical purposes. No easy pattern can be seen. Commented 2 days ago
• Thanks. Please let me know if anything more comes up.
– Max
Commented 2 days ago