Convergence - Infinite Series Can Return a Finite Value
See Carnegie-Mellon's 'Numberous Proofs to the Basel Problem' for Mathematical History and Analytical Solutions
Demonstrated here: the power of calculus in resolving paradoxes created by infinity, and in obtaining analytical solutions to problems that are difficult to approach with numerical analysis alone. The following mathematical concept is convergence, and it applies to continuous functions as well as discrete sums.
Consider the following sequence:
$$\bbox[8px,border:1px solid black] { S_1 = \frac{1}{n^2}\,\ \,\text{where} \,\,n \in \Bbb{N}} \tag{Eq. 1}\label{s1}$$
Expanding this sequence starting at $n=1$ returns a simple list of fractions:
$$\bbox[8px,border:1px solid black] { S_1=\frac{1}{n^2} = 1,\,\frac{1}{4},\,\frac{1}{9},\,\frac{1}{16},\,\frac{1}{25},\,\frac{1}{36},\,...} \tag{Eq. 1.2}\label{1.1}$$
Now let us sum these set elements over the entirety of index n:
$$\bbox[8px,border:1px solid black] { \sum_{n=1}^\infty \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \dots} \tag{Eq. 2}\label{2}$$
On the surface, it looks like the series might sum to infinity regardless of how small these additions become; there are infinitely-many positive additions being made, after all. If adding-together an infinite sequence of numbers can return a finite sum, then what will that number be and how do we even attempt a determination?
- Moreover, these terms belong to the rational number set $\Bbb{Q}$. In finite maths, we cannot sum-together numbers from $\Bbb{Q}$ to obtain a number that is exclusively in $\Bbb{R}$. This is true for all natural numbers $\Bbb{N}$ and integers $\Bbb{Z}$ as well.
Calculus allows us to drop the discrete notation in favor of a continuous function which is analogous to our discrete series -- with domain $x \in [1,\infty)$:
$$\bbox[8px,border:1px solid black]{ f(x)=\int_{1}^{x}\frac{1}{x^2} \ dx} \tag{Eq. 3}\label{3}$$
Below is a plot of $1/x^2$ with its underside shaded to represent the integral $\int{dx/x^2}$. This is our "area-under-the-function" interpration for antiderivatives. This graphical interpretation will resolve a finite area under our infinite curve, but we require a quantitative approach to demonstrate that convergence occurs.
So, we need to integrate $f(x)$ over our domain $[1,\infty)$ and see what analytical solutions - if any - occur.
Since $\infty$ is not itself a value, we should rewrite our improper integral
using limits and a dummy variable $t$, and evaluate:
$$\bbox[8px,border:1px solid black]{ \lim_{t \rightarrow \infty} \int_{1}^{t}\frac{1}{x^2}\ dx = \lim_{t \rightarrow \infty} \left[1 - \frac{1}{t}\right] = 1 - \cancelto{0}{\lim_{t \rightarrow \infty} \left[\frac{1}{t}\right]} = \bbox[#FF1,8px,border:0px solid red]{1}\ } \tag{Eq. 4} $$
Eq. 4 - If $t \rightarrow \infty$, then the denomenator in $\frac{1}{t}$ approaches infinity. Therefore our quotient approaches zero -- and our integral is equal to 1.
Over the domain $[1,\infty)$, $f(x)$ encloses an area of exactly $1$. Determining that this infinite series converges is not trivia; obtaining an analytical solution so easily by applying entry-level calculus is astonishing.
How Can We Demonstrate that this Solution is Not Trivial?
Ask: if $\int_{1}^{\infty} \left[\frac{1}{x^2}\right] dx$ converges to a real and finite value, then will $\int_{1}^{\infty} \left[\frac{1}{x}\right] dx$ also converge? Consider the similarities:
- Both are rational expressions in the form $p/q$
- Monotonically decreasing over $x \in [1,\infty)$
- Has values in-between 0 and 1 over our domain
- Mutual characteristics are graphically demonstrable:
Let us evaluate our continuous sum for $f(x)=\left[\frac{1}{x}\right]$ and see if we diverge or converge:
$$\bbox[8px,border:1px solid black] { \lim_{t \rightarrow \infty} \int_{1}^{t} \left[\frac{1}{x}\right]\ dx = \lim_{t \rightarrow \infty} \ln[t]-\ln[1] = \infty - 0 = \bbox[#FF1,8px,border:0px solid red]{\infty} } \tag{Eq. 5}$$
Despite similarities between the functions, $f_1(x)$ converges over $[1,\infty)$ whereas $f_2(x)$ is divergent.
Evaluating Infinite Series Using Fourier Analysis
Integration of a function over $[1, \infty)$ is far more abstract as a concept than summing a list of fractions.
Therefore, let us return to our discrete sum (Eq. 2) and attempt a solution using Fourier analysis. We can relate our continous function, $f(x)$, to our discrete infinite series $S_1(n)$.
- Generally, Fourier analysis allows us to rewrite any continuous function--with arbitrary precision--as an infinite summation of sines, cosines and coefficients. Convenient for functions that are discontinuous (i.e. electrical signals, environmental vibrations).
- For a function $g(x)$ to have an associated Fourier Series, it must meet the Strong Dirichlet Conditions. Real (physical) signals will always satisfy these conditions! [Citation Required.]
$$\bbox[8px,border:1px solid black] { g(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx)+b_n sin(nx)) } \tag{Eq. 6}$$
The coefficients $a_0$, $a_n$ and $b_n$ are related to $g(x)$ via integration:
$$\bbox[8px,border:1px solid black] { a_0 = \frac{1}{2L} \int_{-L}^{L} g(x) dx } \bbox[8px,border:1px solid black] { a_n = \frac{1}{L} \int_{-L}^{L} g(x) cos\left[n \frac{x}{L}\right]dx } \bbox[8px,border:1px solid black] { b_n = \frac{1}{L} \int_{-L}^{L} g(x) sin\left[n \frac{x}{L}\right]dx } \tag{Eq.s 7-9}$$
Evaluating these three integrals yields one independent coefficient $(a_0)$ and two coefficients $(a_n, b_n)$ which vary with Fourier index $n$. The size of $n$ determines the precision of our series approximation of $g(x)$. Fourier analysis adapts the concept of convergence to fit approximation, computation and analytical purposes.
First introduced to me in Electrodyamics I: Parseval's theorem - valid only for square-integrable $g(x)$ - relates our function to the fourier coefficients as such:$$\bbox[8px,border:1px solid black] { \frac{1}{L} \int_{-L}^{L} \left[g^2(x)\right] dx = \frac{a^2_0}{2} + \sum_{n=1}^{\infty} \left[a^2_n+b^2_n\right] } \tag{Thrm. 1}$$
Given clever choices for $g(x)$, Parseval's theorem simplifies our mathematical approach to determining the value for a converging series.
- An advantage of (Thrm. 1) over (Eq. 6): our Fourier coefficients - $a_n$ and $b_n$ - now dominate their terms in the fourier series of $g(x)$. Integration terms simplify--therefore, our analysis and computational demands do as well.
Let's choose a function and analyze it with Parseval's Theorem to see if we can extrapolate information regarding why - and at what value - $S_1$ will converge.
- Let $g(x)$ = x
- Periodicity: $L= \pi$
$$\bbox[8px,border:1px solid black] { \frac{1}{\pi} \int_{- \pi}^{\pi} x^2\ dx = \frac{a^2_0}{2} + \sum_{n=1}^{\infty} \left[a^2_n+b^2_n\right] } \tag{Eq. 10}$$
Determine Coefficients $a_0$, $a_n$, and $b_n$ for $g(x)=x$
$$ \begin{align*} a_0 &= \frac{1}{\pi} \int_{- \pi}^{\pi} x\ dx\\ &= \frac{1}{\pi} \left[ \frac{x^2}{2} \right]\Bigg|_{-\pi}^{+\pi}\\ &= \frac{1}{\pi} \left[ \frac{\pi^2}{2} - \frac{(- \pi)^2}{2}\right]\\ & \bbox[8px,border:0px solid black] {= \bbox[#FF1,8px,border:0px solid red]{0} } \end{align*} $$
$$ \begin{align*} a_n &= \frac{1}{\pi} \int_{- \pi}^{\pi} x \cos{\left(n x\right)}\ dx\\ &= \frac {1}{\pi} \left[ \frac{\cos{\left(n x\right)}}{n^2} + \require{cancel} \cancelto{0}{\frac{ x \sin{\left(n x\right)}}{n}} \right] \Bigg|_{-\pi}^{+\pi}\\ &= \frac{1}{ \pi} \left[ \frac{ \cos{( - \pi n) } - \cos{( + \pi n) }}{n^2} \right]\\ &= \bbox[8px,border:0px solid black] { \bbox[#FF1,8px,border:0px solid red]{0} } for\ all\ n \end{align*}$$
$$ \begin{align*}
b_n &= \frac{1}{\pi} \int_{- \pi}^{ \pi} x \sin{\left(n x\right)}\ dx\\
&= \frac {1}{\pi}
\left[ \frac{ - \pi \cos{\left(n x\right)}}{n}
+ 0 \right]
\Bigg|_{-\pi}^{+\pi}\\
&= \bbox[8px,border:0px solid black] { \bbox[#FF1,8px,border:0px solid red]{ - \frac{2}{n} \cos{\left(n \pi\right)}} }
\end{align*} $$
Notice that $a_0$ = $a_n$ = 0 in this case because of the integrand's symmetry about the origin; $g(x)=x$ is an odd function: $g(-x)=-g(x)$.
Graphically, it is clear why $a_0 = a_n = 0$ while $b_n \not= 0:$
How a Solution to the Basel Problem Emerges from Parseval's Theorem
Substituting $a_0$, $a_n$, $b_n$ and $g(x)=x$ into (Thrm. 1), we are equating a simple integral to our infinite series:
$$\bbox[8px,border:1px solid black] { \frac{1}{\pi} \int_{- \pi}^{\pi} x^2\ dx = 4 \sum_{n=1}^{\infty} \left[ \frac{ \cos^2{(n \pi)}}{n^2} \right] = \frac{2 \pi^2}{3} } \tag{Eq. 11}$$
Encoded in (Eq. 11): the infinite sum of terms in $\left[cos{(n \pi)}/ n\right]^2$ for $n = (1, 2, 3, 4... \infty)$ is equal to a finite number - exactly $ \frac{ \pi^2}{6}$! Parseval's Theorem allowed us to simply integrate $ \int x^2dx$ to reach our conclusion.
- Congratulations! We have proven that certain infinite series will converge to a finite number.
- Possible because $\int x^2 dx$ is easily obtainable in comparison to $ \sum_{n=1}^{ \infty} \frac{cos^2(n)}{n^2}$
- We can take this a step further to actually solve the Basel Problem (Eq. 2) by analyzing $cos(n \pi)$ a little closer. Consider a table of cosine outputs for different $n$:
Cosine has a period of only $2 \pi$, so this table contains our only two unique outputs of $ \cos(n \pi)$:
Incredibly, the result is an irrational number! It is not trivial at this point that $\pi$ should appear. Why?
- Every term in our series belongs to the rational number set $\Bbb{Q}$, but our sum converges outside of that set to an irrational number in $\Bbb{R}$!
Leonhard Euler, (1707-1783)
Note that convergence was a known property of infinite series prior to Euler.
His discovery (in this context) is solving the Basel Problem.