1
$\begingroup$

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.

$\endgroup$
1
  • 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

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ $(+1)$. 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
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.