# Stirling approximation for factorials

## [Part 1] Testing the convergence of series.

When evaluating distribution functions for statistics, it is often necessary to evaluate the factorials of sizable numbers. Stirling formula is an approximation for factorials that leads to a very accurate results of the factorials. In this series of posts, we will try to get a mathematical intuition of this formula.

In this first part we will try to have a closer look at the behavior of infinite series and some of their properties. The motivation behind is to have a better intuition about some very interesting concepts and tools used in everyday mathematics, physics and computer science.

## Definition

We call a series associated with sequence ${\left(U_n\right)}_{n \in {\mathbb{N}} }$ and noted $\sum U_n$ or $\sum_{n \in {\mathbb{N}}} U_n$, the couple:

where $\forall n \in {\mathbb{N}} , S_n = \sum_{k=0}^{n} U_k$

$S_n$ is called the partial sum of the series $\sum U_n$

## Convergence and divergence

Let $\left( {\mathbb{E}}, \|.\| \right)$ be normed vector space, $\sum U_n$ a series defined in ${\mathbb{E}}$ and ${\left( S_n \right)}_n$ the sequence of partial sums.

• We say that the series $\sum U_n$ converges (respectively diverges) in $\left( {\mathbb{E}}, \|.\| \right)$ if the sequence ${\left( S_n \right)}_n$ converges (respectively diverges) in $\left( {\mathbb{E}}, \|.\| \right)$.
• In this case, the limit $S$ of the sequence ${\left( S_n \right)}_n$ is called the sum of the series $\sum U_n$, and is noted: $\sum_{n=0}^{\infty} U_n = S = \lim \limits_{n \to \infty} S_n$

Examples

• the geometric series for complex numbers.

We have the following, $\forall n \in {\mathbb{N}}$

Meaning that the series converges iff $\|q\| \lt 1$

and in this case $\sum_{n = 0}^{+\infty} q^n = \frac{1}{1-q}$

The result is simple yet so impressive, a lot of philosophers struggled to resolve Zeno’s paradoxes because they lacked the mathematical tools to correctly model the problem.

• another example $\sum_{n \gt 0} \frac{1}{n(n+1)} , \forall n \in {\mathbb{ N^* }}$

Hence the series $\sum\frac{1}{n(n+1)}$ converges, and the infinite sum is 1.

## Convergence tests

### 1- The n-th term test (convergence’s necessary condition)

Let $\left( {\mathbb{E}}, \|.\| \right)$ be normed vector space, $\sum U_n$ a series defined in ${\mathbb{E}}$

Proof

Let $S_n = \sum_{k = 0}^{n}$ be the partial sums of the series, we know that the series converges if the $\lim \limits_{n \to \infty = S}$

Hence, we have $U_n = S_n - S_{n-1} \require{AMScd} \begin{CD} @>>{n \to \infty}>\end{CD} 0$

N.B. 1

A direct conclusion from this test is that, if the general term $U_n$ does not converge to 0, then the series diverges.

Again if we look at the geometric series which converges only if $\|q\| \lt 1$, we can see that if $\|q\| >= 1, \lim \limits_{n \to \infty} q^n > 0$

N.B. 2

having the general term $U_n$ converging to 0, does not guarantee the convergence of the series.

Let’s study the series $\sum log \left( \frac{n+1}{n} \right)$

We have the following $\lim \limits_{n \to \infty} log \left( \frac{n+1}{n} \right) = 0$

Hence the series diverges.

### 2- Cauchy’s condition

Let $\left( {\mathbb{E}}, \|.\| \right)$ be normed vector space, $\sum U_n$ a series defined in ${\mathbb{E}}$

Proof

The proof is quite simple if we consider the sequence of partial sums ${(S_n)}_{n}$. The series converges implies that $(S_n)_{n}$ converges, which means that it’s a Cauchy sequence.

Example

Let’s consider the harmonic series $\sum \frac{1}{n}$

Hence the series does not verify the Cauchy’s condition, and so it does not converge.

N.B.

This condition is important, and most of the time is used to study series in Banach spaces.

Let ${\mathbb{E}}$ be a Banach space:

• A series converges $iff$ it verifies the Cauchy’s condition.
• A series converges absolutely $\Rightarrow$ it converges.

In the second property the convergence does not imply the absolute convergence, as an example let’s consider the following series $\sum \frac{ (-1)^{n+1} }{n}$

And since $\lvert \int_{0}^{1} \frac{ (-x)^n }{ 1 + x } dx \lvert \le \int_{0}^{1} (-x)^n = \frac{ 1 }{ n + 1 } \require{AMScd} \begin{CD} @>>{n \to \infty}>\end{CD} 0$

It follows that $\lim \limits_{ n \to \infty } S_n = ln(2)$

Hence the series converges to $ln(2)$ but it does not converge absolutely.

Let’s check this result in python, we will use the symbolic library sympy

 1 from sympy import symbols, Sum, oo
2
3 k, n = symbols('k n', integer=True)
4 ps = Sum((-1)**(k+1)/k, (k, 1, n))
5
6 #  lets calculate the partial sum
7 ps.doit()
8 > (-1)**(n + 1)*lerchphi(exp_polar(I*pi), 1, n + 1) + log(2)
9
10 #  the infinite sum
11 s = ps.subs(n, oo)
12 s.doit()
13 > log(2)
14 s.evalf()
15 > 0.693147180559945

N.B.

A series that converges but does not converge absolutely, is called semi-convergent or conditionally convergent.

### 3- Leibnitz rule for testing the convergence of alternating series

Let $\sum U_n$ be a series over reals, if $\sum U_{n}$ is an alternating series, i.e. $U_n = (-1)^{n} \lvert U_n \lvert$ or $U_n = (-1)^{ n + 1 } \lvert U_n \lvert$, and $(\lvert U_n \lvert)_n$ is decreasing and converges to 0, then the series $\sum U_n$ converges and $R_n = S - S_n \lt \lvert U_{ n + 1} \lvert$

Proof

Let’s suppose that $\forall n \in {\mathbb{ N }} U_n = (-1)^n \lvert U_n \lvert$ and $\forall n \in {\mathbb{ N }} \,\, S_n = \sum_{k=0}^{n} U_k$

Hence ${(S_{2n})}_n$ is decreasing. The same way we can show that ${(S_{2n + 1})}_n$ is increasing. And since ${(S_{2n + 1})}_n - {(S_{2n})}_n = U_{2n + 1} \require{AMScd} \begin{CD} @>>{n \to \infty}>\end{CD} 0$, both ${(S_{2n + 1})}_n$ and ${(S_{2n})}_n$ converge to the same limit $S$, then we find that $\lim \limits_{n \to \infty} S_n = S$.

Also, since

Example

### 4- Comparison test

Proof

We know that $S_{n + 1} - S_n = U_{n + 1} >= 0$ and that the sequence ${(S_n)}_n$ is increasing.

Based on this property we can deduce many other properties related to comparing series in different ways, i.e. using $\sim , \, o , \, O , \, ...$

Example: Riemann series

We cal a Riemann series, a series of the form $\sum \frac{1}{ n^\alpha } , \, \alpha \in {\mathbb{R}}$

• Case when $\alpha \le 0$

The series $\sum \frac{1}{ n^\alpha }$ obviously diverges.

• Case when $\alpha = 1$

We saw that $\sum \frac{1}{n}$ diverges.

• Case when $\alpha \in [ 0, 1 ]$

The series $\sum \frac{1}{ n^\alpha }$ diverges, because $\frac{1}{n} =o \frac{1}{ n^\alpha }$

• Case when $\alpha > 1$

Let’s consider the application $f$

The application $f$ is of class $C^1$, and by applying the Mean value theorem over $[n, n+1]$, we have the following result $\exists \theta_n \in [n, n+1]$ such that $f\left( n+1 \right) - f\left( n \right) = f^{''}\left( \theta_n \right) = \frac{ 1 }{ { \theta_n }^\alpha }$

And since $\theta_n \sim n$

It follows that $\frac{1}{ { \theta_n }^\alpha } \sim \frac{1}{ n^\alpha }$

Which means that $\frac{1}{ n^\alpha } \sim \left( f\left( n+1 \right) - f\left( n \right) \right)$

We know that the series $\sum \left( f\left( n+1 \right) - f\left( n \right) \right)$ converges, because $\sum_{ k = 1 }^{ n } \left( f\left( n+1 \right) - f\left( n \right) \right) = f\left( n+1 \right) - f\left( 1 \right) \require{AMScd} \begin{CD} @>>{n \to \infty}>\end{CD} -f \left( 1 \right)$

Finally $\sum \frac{1}{n^\alpha}$ converges if $\alpha > 1$.

 1 from sympy import symbols, Sum, oo
2
3 a = symbols('a')
4 n, k  = symbols('n k', integer=True)
5 ps = Sum(1/(k**a), (k, 1, n))
6 s = ps.subs(n, oo)
7
8 #  the infinite sum for a < 0
9 s.subs(a, -2).doit()
10 > oo
11
12 #  the infinite sum for a == 1
13 s.subs(a, 1).doit()
14 > oo
15
16 #  the infinite sum for a > 1
17 s.subs(a, 2).doit()
18 > pi**2/6
19
20 s.subs(a, 3).doit()
21 > zeta(3)
22
23 s.subs(a, 4).doit()
24 > pi**4/90
25
26 s.subs(a, 5).doit()
27 > zeta(5)