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:


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.

  • 4
    $\begingroup$ I'd recommend asking this question on the math SE because it does not seem suited for Mathematica analysis. $\endgroup$
    – Roman
    Commented 2 days ago

2 Answers 2


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.

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

discrete plot of the sequence f_n

The numbers seem to be pretty random and spread uniformly over $[0,1]$:

Histogram[Array[g, 1000], {0, 1, 1/10}]

histogram of the first 1000 values

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.

  • $\begingroup$ Thanks. ($+1$). How to show that the sequence $f_n$ is quasi-random and uniform on $[0,1]$? $\endgroup$
    – Max
    Commented 2 days ago
  • $\begingroup$ 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. $\endgroup$
    – Roman
    Commented 2 days ago
  • $\begingroup$ What do you mean by quasi- random and uniform $\endgroup$
    – Max
    Commented 2 days ago
  • $\begingroup$ I mean a low-discrepancy sequence. Quasi-random, for practical purposes. No easy pattern can be seen. $\endgroup$
    – Roman
    Commented 2 days ago
  • $\begingroup$ Thanks. Please let me know if anything more comes up. $\endgroup$
    – Max
    Commented 2 days ago

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