Multi-Factorial and Series with Triple-factorial terms

Question

Let $n!^{(k)}$ denote a multi-factorial which is defined by $$ n!^{(k)} = \begin{cases} 1 & n \leqslant 0, \\ n, & 0 < n < k,\\ n\times(n-k)!^{(k)}, & n \geqslant k. \end{cases} $$

E.g. $8!^{(3)}=8\times5!^{(3)}=8\times5\times2!^{(3)}=8\times5\times2=80$. For $k=3$ we may also put down $8!!!$.

For the case of $k=1$, Mathematica has Factorial[] or ! at hand.

For the case of $k=2$, there is Factorial2[] or !! which calculates double factorials.

For the general case of a $k$^th factorial, I could define by recursion

In[1]:= MultiFactorial[n_, k_] := If[n < k, If[n <= 0, 1, n], n MultiFactorial[n - k, k]]

which correctly outputs $8!^{(3)}$ as

In[2]:= MultiFactorial[8, 3]
Out[2]:= 80

But when I have a multi-factorial in every term of an infinite series, Mathematica spits out the original input:

In[3]:= Sum[((-1)^n MultiFactorial[3 n, 3])/MultiFactorial[3 n + 1, 3], {n, 0, +∞}]
Out[3]:= (* More or less the same *)

[Edit]

Definition by Product[]

In[4]:= MultiFactorial[n_, k_] := Product[i, {i, n, 1, -k}]
In[5]:= Sum[((-1)^n MultiFactorial[3 n, 3])/MultiFactorial[3 n + 1, 3], {n, 0, +∞}]

suggests that the sum is divergent.

Yep, I'm trying to verify if

$$\sum\limits_{n=0}^{+\infty} \frac{(-1)^n (3n)!^{(3)}}{(3n+1)!^{(3)}} = -\frac{\sqrt[3]{2}}{4} \ln\left(\sqrt[3]{2}-1\right)+\frac{\sqrt{3}\sqrt[3]{2}}{6}\arctan\frac{\sqrt{3}}{1+2\sqrt[3]{2}}.$$

Question: Are there any built-in functions for the cases of $k\geqslant3$ which is OK to sum?

Answer 1

(too long for a comment)

Sasha has given an expression in terms of falling factorials. For some reason, however, Sum[(-1)^j MultiFactorial[3 j, 3]/MultiFactorial[3 j + 1, 3], {j, 0, Infinity}] remains unsimplified using the definition of the multifactorial in terms of FactorialPower[].

Either of the following definitions work, though:

MultiFactorial[n_, k_] :=
      With[{q = Quotient[n - 1, k]}, k^(q + 1) Gamma[n/k + 1]/Gamma[n/k - q]]

MultiFactorial[n_, k_] := With[{q = Quotient[n + k - 1, k]}, k^q q! Binomial[n/k, q]]

where Sum[(-1)^j MultiFactorial[3 j, 3]/MultiFactorial[3 j + 1, 3], {j, 0, Infinity}] now evaluates to Hypergeometric2F1[1, 1, 4/3, -1]. Applying FunctionExpand[] to this yields a rather complex expression, but at least it involves only elementary functions.

Unfortunately,

FullSimplify[FunctionExpand[
              Hypergeometric2F1[1, 1, 4/3, -1] ==
              -2^(-5/3) Log[2^(1/3) - 1] + Sqrt[3] 2^(1/3) ArcTan[Sqrt[3]/(1 + 2^(4/3))]/6]]

takes quite long to evaluate, and does not yield anything useful. Thus, we need to exploit a few special function identities like the Pfaff transformation, yielding the identity

$${}_2 F_1\left({{1,1}\atop{\frac43}}\mid-1\right)=\frac12{}_2 F_1\left({{\frac13,1}\atop{\frac43}}\mid\frac12\right)$$

which happens to have a convenient representation in terms of elementary functions.

Using that identity, we have

With[{z = 1/2}, (-Log[1 - z^(1/3)] - (-1)^(2/3) Log[1 + (-1)^(1/3) z^(1/3)] +
     (-1)^(1/3) Log[1 - (-1)^(2/3) z^(1/3)])/(3 z^(1/3))]/2 ==
-2^(-5/3) Log[2^(1/3) - 1] + Sqrt[3] 2^(1/3) ArcTan[Sqrt[3]/(1 + 2^(4/3))]/6 // FullSimplify

which quickly evaluates to True.

My point here, now, is that although Mathematica knows a fair bit about special functions, sometimes one could benefit a lot from helping Mathematica a little bit with a few identities...

Answer 2

The multifactorial you are building is built-in in Mathematica under the disguise of FactorialPower:

MultiFactorial2[n_, k_] := FactorialPower[n, Quotient[n - 1, k] + 1, k]

which is the equivalent to J. M.'s expression.

Answer 3

Mathematica actually can perform the sum as originally specified. First, supply the recursive definition of "multi-factorial":

factorial[n_, k_] /; n <= 0 := 1;
factorial[n_, k_] /; n <= k := n;
factorial[n_, k_] := factorial[n, k] = n factorial[n - k, k];

The strategy is to simplify the coefficients that will appear in the sum. Unfortunately, a direct attack won't work:

a[n_] := Evaluate[Simplify[factorial[3 n, 3] / factorial[3 n + 1, 3], 
  Assumptions -> n \[Element] Integers && n >= 0]]

This doesn't get us anywhere. Instead, let's try...guessing:

a = FindSequenceFunction[Table[factorial[3 n, 3] / factorial[3 n + 1, 3], {n, 1, 10}]]

$\frac{\text{Pochhammer}[1,\text{$\#$1}]}{\text{Pochhammer}\left[\frac{4}{3},\text{$\#$1}\right]}\&$

We are home free. We can even do a more general sum, using a variable x instead of $-1$:

f[x_] := Evaluate[Sum[x^n a[n], {n, 0, Infinity}] // FullSimplify]

The answer:

f[-1]

$\text{Hypergeometric2F1}\left[1,1,\frac{4}{3},-1\right]$

(Its connection with ArcTan and Log becomes apparent after looking at FunctionExpand[f[x]].)

---

One should be sceptical of the result of FindSequenceFunction--it is only guaranteed to agree with the sequence of values given to it, not with all possible indexes--so let's find a way to confirm the correctness of this result. All we have to go on is the recursive definition of factorial. Somehow we need to relate a to this recursion. Looking at a shows it relates factorial[3n,3] to factorial[3n+1,3], whereas the recursive definition of factorial relates factorial[3n,3] to factorial[3n+3,3]. We can get there in three steps, but first need to work out the next two:

b = FindSequenceFunction[Table[factorial[3 n - 1, 3]/factorial[3 n, 3], {n, 1, 10}]]
c = FindSequenceFunction[Table[factorial[3 n - 2, 3]/factorial[3 n - 1, 3], {n, 1, 10}]]

Now 1/(a[n-1]b[n]c[n]) ought to give us the ratio factorial[3n]/factorial[3(n-1)], which, according to the recursion, had better equal $3n$. Indeed,

1/(a[n - 1] b[n] c[n]) // FullSimplify

$3n$

Because the first four values of factorial are $(1,1,2,3)$, we only need to verify that a[0], c[1], and b[1] return the first three successive ratios $1/1$, $1/2$, $2/3$, which indeed they do. By induction, this proves the result of FindSequenceFunction is correct.