Sequences and Series of Factorials

A Lesson by cyphertext

This is my original work on a combinatorial problem. It is advised that you understand advanced calculus or calculus 3 to read it.

Factorial Sequences and Series

By: Steven Tao Zheng

23/03/2011

Print Notes: Sorry for the bad layout of equations, the original equations editor tool on Microsoft Word did not show when I pasted my document. If you want a copy with better quality, email me at [email protected].

Legend

_a^b▒ = from limits a to b (of an integral or sum)

〖〗= type of brackets

lim┬(n→∞)⁡= the limit as n approaches infinity

t_n = t subscript n

Abstract

Arithmetic sequences and series deal with the growth and decay of patterns that have a common difference. Geometric sequences and series deal with the growth and decay of patterns that have a common rate. This introduction to factorial sequences and series is a definitive collection of theories and theorems that deal with a structured but unstable growth through the implementation of analytical reasoning. The purpose of this study is to derive a simple formula using analysis to build a standing theory on factorial sequences and series.

Introduction

Consider the following problem: 1! +2! +3! +⋯+N!

To this date, this simple yet elusive combinatorial problem has yet provided a simple formula for evaluation (a formula without the usage of special non-elementary functions).

Theory

Axioms

1)The value of n and its factorial are natural numbers.

2)For factorials n≤6, the order of magnitude might be shared by more than one value of n.

3)The order of magnitude for factorials where n is greater than six are distinct, meaning they are different exponentiations of ten.

4)For every value of n, there is only one value for n!.

1.0 Sequence of Factorials

Mathematical sequences are sets of numbers whose behaviour follows a set pattern. When dealing with factorial sequences, only the natural numbers play a role, making the entire sequence discrete. [Axiom 1]

Let t_n describe a factorial sequence.

t_n t_n=k f (t_n) =c (t_n)

t_1 1 f (t_1) =c

t_2 2 f (t_2) =2c

t_3 6 f (t_3) =6c

t_n n! f (t_n )=n!c

1.1 Finding the nth term of a factorial sequence

Suppose a problem involving finding the nth term of a factorial sequence looked like this:

(f (t_n))/ (f (t_3)) =20

f (t_n )=20*3!c

f (t_n )=5!c

n!c=5!c

n!=5!

n=5

Small factorials can be easily determined.

However for a problem like this one:

(f (t_n))/ (f (t_7)) =2.413593262×〖10〗^13

(n=19)

The problem presented above is difficult to solve mentally.

1.2 Finding the quotient of two terms of sequence

(f(t_18))/(f(t_5))= 18!c/5!c=5.335311421×〖10〗^13

Conceptually, the nth terms’ are identical to the value of n and the constants cancel.

1.3 Stirling’s Approximation and Augmentations

Stirling’s Approximation states lim┬(n→∞)⁡〖((√2π)*n^(n+1) *e^(-n))/n!〗 = 1

Therefore n!≈ (√2π)*n^(n+1)* e^(-n)

Despite the Stirling’s Approximations asymptotic approach, the accuracy is not precise. However, the accuracy increases as the value of n increases.

1.4 Analytical Refinement of the Sterling’s Approximation

From the famous Stirling’s approximation, we can deduce that:

(√2π)*n^(n+0.5)* e^(-n)*√2π)*n^(n+1) e^(-n)

However, we want an approximation that is precise to at least three decimals for all values greater than n=14. Since the Stirling’s Approximation increases in precision as n increases, so does any proper augmentation.

(√2π)* n^(n+0.5)* e^(-n)*k≈n! would lead to a closer approximation where 1

The value of k should asymptotically approach 1 as n increases as well as maintaining at least three decimals of precision for n factorial beyond n=14.

Let k= e^c

If lim┬(c→∞)⁡〖e^c 〗=1

Then c=1/(m(n)+q) , where m and q are constants.

Then e^(1/12n) is close to the precision range

Verification

5!=120

5!≈(√2π)*5^(5+0.5) *e^(-5) *e^(1/5m)

5!≈118.019168e^(1/5m)

m=11 5! = 120.2584396 δ=+0.2584396

m=12 5! = 120.0026371 δ=+0.0026371

m=13 5! = 119.8488862 δ= -0.1511138

12 is a good approximate for the coefficient m.

e^(1/12n)

5!≈(√2π)*5^(5+0.5) *e^(-5)* e^(1/(60+b))

δ=n!-[(√2π)* n^(n+0.5) *e^(-5) *e^(1/(12n+b))]

b=0.1 5! = 119.9993093 δ= -6.90685×〖10〗^(-4)

b=0.5 5! = 119.9861089 δ= -0.0138911

b=0.9 5! = 119.9730835 δ= -0.0269165

20!≈(√2π)〖20〗^(20+0.5) e^(-20) e^(1/(240+b))

20!=2.432902008×〖10〗^18

b=0.1 20! = 2.432898630×〖10〗^18 δ=+3.378×〖10〗^12

b=0.5 20! = 2.432881777×〖10〗^18 δ=+2.0231×〖10〗^13

Therefore e^c≈e^(1⁄((12n+0.1) ))

Refined Stirling’s Approximation

n!≈(√2π)*n^(n+0.5) *e^(-n) *e^(1/(12n+0.1))

1.5 Solving sequences with increased precision

n!≈(√2π)n^(n+0.5) e^(-n) e^(1/(12n+0.1))

(f(t_n))/(f(t_p))=c Solve for f(t_n ) when f(t_p ) is known.

f(t_n )=c*f(t_p)

Let k=c*f(t_n)/(√2π)

(f(t_n))/(f(t_p))=c Solve for f(t_p ) when f(t_n ) is known.

f(t_p )=1/c*f(t_n)

Let k=f(t_n)/(c*√2π)

Let f(t_n )≈〖(n/e)〗^n since the value of e^(1⁄(12n+0.1 ))decreases as n increases, making it ultimately negligible.

ln⁡〖〖(n/e)〗^n 〗=ln⁡k

n[ln⁡(n)-ln⁡(e)]=ln⁡k

n[ln(n)-1]=ln⁡ k

n[ln⁡(n)-1]< ln⁡ k<(n+1)(ln⁡(n+1)-1)

Therefore (n+0.5)((ln (n+0.5))-1)≈ln⁡ k

Using Newton’s Method, the value of n can be solved.

f(n)=(n+0.5)((ln⁡(n+0.5))-1)-ln⁡ k

f^' (n)=ln⁡(n+0.5)

This class of function may implement the zero as an initial estimate.

x_1=|0-f(0)/(f^' (0) )|

x_2=|x_1-f(x_1 )/(f^' (x_1 ) )|

x_3=|x_2-f(x_2 )/(f^' (x_2 ) )|

x_n=|x_(n-1)-f(x_(n-1) )/(f^' (x_(n-1) ) )|

1.6 The Concept of One to One Correspondence for above methods

The methods above for approximating the values of factorials and solving sequences have one common property: one to one correspondence. For every value n, there is only one distinct value of n factorial. The refined approximations are designed to be precise to the point the one to one correspondence is clear. [Axiom 4]

Factorial Series

Factorial series are sums of consecutive factorials as dictated by the factorial sequence. This formula is derived using Gamma functions.

Equation 1

n!+(n+1)!+(n+2)!+(n+3)!+⋯+(n+p)!=∑_(k=n)^(n+p)▒k!

Equation 2

∑_(k=n)^(n+p)▒k!=Г(n+1)+Г(n+2)+Г(n+3)+Г(n+4)+⋯+Г(n+p+1)

Equation 3

Г(n+1)+Г(n+2)+Г(n+3)+Г(n+4)+⋯+Г(n+p+1)=∫_0^∞▒〖e^(-t) t^n 〗 dt+∫_0^∞▒〖e^(-t) t^(n+1) 〗 dt+∫_0^∞▒〖e^(-t) t^(n+2) 〗 dt+∫_0^∞▒〖e^(-t) t^(n+3) 〗 dt+⋯+∫_0^∞▒〖e^(-t) t^(n+p) 〗 dt

Equation 4

∫_0^∞▒〖〖(e〗^(-t) t^n+e^(-t) t^(n+1) 〗+e^(-t) t^(n+2)+e^(-t) t^(n+3)+⋯+e^(-t) t^(n+p))dt=∑_(k=n)^(n+p)▒k!

Equation 5

∫_0^∞▒〖〖e^(-t) (t〗^n+t^(n+1) 〗+t^(n+2)+t^(n+3)+⋯+t^(n+p))dt=∑_(k=n)^(n+p)▒k!

Equation 6

∫_0^∞▒〖e^(-t) t^n ((t^(p+1)-1)/(t-1))〗 dt=∑_(k=n)^(n+p)▒k!

Equation 7 (Final)

∫_0^∞▒(t^(n+p+1)-t^n)/(e^t (t-1)) dt=∑_(k=n)^(n+p)▒k!

Consequently,

∫_0^∞▒(t^n-t^(n+p+1))/(e^t (t-1)) dt=∑_(k=n)^(n+p)▒〖-(k!)〗

Factorial Series Theorem or Factorial Sum Function

c∫_0^∞▒(t^(n+p+1)-t^n)/(e^t (t-1)) dt=∑_(k=n)^(n+p)▒〖c(k!)〗

The variables of change in the summation of consecutive factorials are c, n and p, where c is a constant, n is the initial term and p is the difference of terms from n.

2.1 Solving Factorial Series

Finding the value of a factorial series where the values c, n and p are known.

∑_(k=n)^(n+p)▒〖c(k!)〗=Q

If the numbers are small for n and p, then manual calculations can be easily executed. For larger numbers, the values of c, n and p should be directly substituted to the Factorial Series Theorem and possibly using numerical methods of integration to compute its true value.

Solving for c when Q and ∑_(k=n)^(n+p)▒k! Are known.

∑_(k=n)^(n+p)▒〖c(k!)〗=Q

c∑_(k=n)^(n+p)▒k!=Q

c=Q/(∑_(k=n)^(n+p)▒k!)

When consecutive factorials are summed, the largest term in the series is the easiest to solve for and the smallest term is the hardest to solve for because the order of magnitude for the largest term is responsible for most of the series’ make-up, whereas the smaller terms may be negligible.

Solving for n+p when Q and c are known

∑_(k=n)^(n+p)▒c(k!) =cQ

∑_(k=n)^(n+p)▒k!=Q

The order of magnitude of Q should be closest to if not identical with (n+p)!. Therefore, we can solve for n+p, with Newton’s method introduced in section 1.5.

2.2 Factorial Difference Series

Factorial difference series are differences of consecutive factorials as dictated by the factorial sequence. This type of problem is considered a combinatorial decay problem.

Equation 1

n!-(n+1)!-(n+2)!-(n+3)!-…-(n+p)!=n!-∑_(k=n)^(n+p-1)▒〖(k+1)!〗

Equation 2

n!-∑_(k=n)^(n+p)▒(k+1)!=Г(n+1)-Г(n+2)-Г(n+3)-Г(n+4)-…-Г(n+p+1)

Equation 3

∫_0^∞▒〖〖e^(-t) (t〗^n-t^(n+1) 〗-t^(n+2)-t^(n+3)-…-t^(n+p))dt=n!-∑_(k=n)^(n+p-1)▒(k+1)!

a(1-r^2-r^3-…-r^n )=a(1-(r+r^2+r^3+⋯+r^n-1))

a(1-r^2-r^3-…-r^n )=a(1-((r^n-1)/(r-1)-1)

Therefore,a(1-r^2-r^3-…-r^n )=a(2-(r^n-1)/(r-1))

Equation 4 (Final)

∫_0^∞▒〖e^(-t) t^n (2-(t^(p+1)-1)/(t-1))〗 dt=n!-∑_(k=n)^(n+p-1)▒(k+1)!

Factorial Difference Series Theorem

c∫_0^∞▒〖e^(-t) t^n (2-(t^(p+1)-1)/(t-1))〗 dt=c(n!-∑_(k=n)^(n+p)▒(k+1)!)

Where c is a constant.

2.3 Additional Theorems

n!-(n-1)!-(n-2)!-(n-3)!-…-(n-p)!=n!-∑_(k=n)^(n-p+1)▒〖(k-1)!〗

c∫_0^∞▒〖e^(-t) t^n (2-(〖(1/t)〗^(p+1)-1)/((1/t)-1))〗 dt=c(n!-∑_(k=n)^(n-p+1)▒(k-1)!)

This theorem is based on the fact:

a(r^n-r^(n-1)-r^(n-2)-…-r^1 )=a(1-((1/r)+(1/r)^2+(1/r)^3+⋯+(1/r)^n-1))

Therefore,a(r^n-r^(n-1)-r^(n-2)-…-r^1 )=a(2-((1/r)^n-1)/((1/r)-1))

3.1Arithmetic Method for Evaluating Large Factorial Sums Part 1

Integration of the Factorial Sum Function

Although the formulas above are simple compared to the formulas of their predecessors, they are still considered implicit, simply because the formulas are expressed as improper integrals. The validity and precision of these formulas are total, but the formulas fail to explicitly give numerical results.

∫_0^∞▒(t^(n+p+1)-t^n)/(e^t (t-1)) dt=∑_(k=n)^(n+p)▒k!

This expression can be written as

∫_0^∞▒〖e^(-t) t^n (1+t+t^2+t^3+⋯+t^p)〗 dt=∑_(k=n)^(n+p)▒k!

Using integration by parts:

Let u=t^n+t^(n+1)+t^(n+2)+⋯+t^(n+p)

du=[nt^(n-1)+(n+1) t^n+(n+2) t^(n+1)+⋯+(n+p) t^(n+p) ]dt

dv=e^(-t) dt

v=-e^(-t)

Suppose the objective is to find the indefinite integral of this expression, the employment of successive integration by parts will be required.

∫▒〖udv=uv-∫▒vdu〗

∫▒〖e^(-t) t^n (1+t+t^2+t^3+⋯+t^p)dt〗= -e^(-t) (t^n+t^(n+1)+t^(n+2)+⋯+t^(n+p) )+ ∫▒[nt^(n-1)+(n+1) t^n+(n+2) t^(n+1)+⋯+(n+p) t^(n+p) ]dt

This method terminates by the p+1 derivative. Hence in the end, the integral becomes:

∫▒〖e^(-t) t^n (1+t+t^2+t^3+⋯+t^p )dt〗= -e^(-t) (u+u^'+u^''+⋯+u^(n+p+1)')

Arithmetically speaking, the value of,

∑_(k=n)^(n+p)▒k!=-e^(-t) (∑_(x=0)^(x=n+p+1)▒〖u^((x)') 〗)

Example

∫_0^∞▒(t^5-t^2)/(e^t (t-1)) dt=2!+3!+4!=32

∫▒〖(t^5-t^2)/(e^t (t-1)) dt= -e^(-t) (〗 t^4+5t^3+16t^2+32t+32)

This implies that the value of ∫_0^∞▒(t^5-t^2)/(e^t (t-1)) dt is equal to the absolute value of the y-intercept of ∫▒〖(t^5-t^2)/(e^t (t-1)) dt〗.

3.2 Arithmetic Method for Evaluating Large Factorial Sums Part 2

The Ascent Base Notation Part 2(An arithmetic convention)

Base Two

In binary (base two), the number 1101 is equal to 13 in base ten.

1101=1(2^3 )+1(2^2 )+0(2^1 )+1(2^0 )=13

There exists a shortcut to the evaluation of base two.

(((((1*2)+1)*2)+0)*2)+1=13

Ascent Base

Notation

1_(n+p)…1_(n+2) 1_(n+1) 0_n 0_(n-1)…0_etc

Examples

5!+4!+3!+2!+1!=1_5 1_4 1_3 1_2 1_1=153

(((((((((1*5)+1)*4)+1)*3)+1)*2)+1)*1)+1=153

6!+5!+4!=1_6 1_5 0_4 0_3 0_2 0_1=864

((((1*6)+1)*5)+1)*4!=864

Significance

By transforming the sum of consecutive factorials into the consecutive sums of Gamma Functions, it is evident that there exists a link between factorials and geometric progression.

Sources

Wrede, Robert C. and Spiegel Murray. Schaum’s Outlines Advanced Calculus. McGraw-Hill Companies, Inc. Second Edition, 2002.

Credible Online Sources

Factorial Sums. Wolfram Mathworld. 24 Feb. 2011