Infinite sum done right
Tianran Chen
Department of Mathematics
Auburn University at Montgomery
Today, we aim to gain a clear understanding of "series".
A sequence is simply a (possibly infinite) ordered the list of numbers. We are mainly interested in sequence of real numbers. E.g., $1,2,3,4,\dots$.
We use notations $( a_n )_{n=1}^\infty$, or simply $( a_n )$ (or $\{ a_n \}$) together with the formula for each $a_n$. (Not to be confused with sets)
For example, the sequence of all positive odd numbers can be represent by the sequence $( b_k )_{k=0}^{\infty}$ given by \[ b_k = 2k + 1. \]
Alternatively, we also consider sequences as functions that maps integers to real numbers. E.g., the function \[ f(n) = 2n + 1, \] when restricted to the domain of nonnegative integers, also produce the sequence of odd numbers.
In general, the sequence $(a_n)_{n=1}^\infty$ is equivalent to the function \[ f(n) = a_n \] that is defined on the set of natural numbers.
The most important operation is computing limits of sequences.
Definition. We say $L$ is the limit of a sequence $(a_n)$ (or $(a_n)$ converges to $L$) if for any $\epsilon > 0$, there exists a natural number $N$ such that \begin{align*} n &\ge N &&\text{implies} & |a_n - L | &< \epsilon. \end{align*} In this case, we say the sequence $(a_n)$ is convergent. Otherwise, we say the sequence is divergent.
The skill and intuitions about limits of functions you acquired still work here (e.g. the squeeze/sandwich theorem).
A series is a "formal" sum of infinitely many objects. For the series with the terms $a_1,a_2,\dots$, we use both notations \[ \sum_{k=1}^\infty a_k = a_1 + a_2 + a_3 + a_4 + \cdots \]
For example, the series \[ \sum_{k=1}^\infty \frac{1}{2^k} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \] is an infinite sum of all positive integral powers of $\frac{1}{2}$.
We should not take the concept of infinite sum for granted.
Consider the series \[ 1 + (-1) + 1 + (-1) + 1 + (-1) + \cdots \]
We can group the consecutive terms and get \[ [1 + (-1)] + [1 + (-1)] + [1 + (-1)] + \cdots = 0 + 0 + 0 + \cdots \] which must be 0 under any reasonable interpretation.
However, the same series can be grouped in a different way: \[ 1 + [(-1) + 1] + [(-1) + 1] + \cdots = 1 + 0 + 0 + \cdots \] which appears to be 1. This does not make sense!
If we add up infinitely many terms together, shouldn't we just get $\infty$?
Consider the series \begin{align*} 1 + 1 + 1 &+ \cdots &&\text{or} & 1 + 2 + 3 &+ \cdots \end{align*} There is no reason to think they can add up to any finite value.
But the series \[ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \] will indeed produce a finite value. Can you see what should that value be?
To distinguish the kind of series that actually make sense and add up to finite value from the rest, we use the concept of "partial sums".
For a series $\sum_{k=1}^\infty a_k$, we define its $n$-th partial sum to be $ S_n = \sum_{k=1}^n a_k $, i.e., the sum of the first $n$ terms. If \[ L = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k=1}^n a_k \] exists (and finite), then this series converges to $L$ (is convergent). Otherwise, this series diverges. (is divergent).
The "partial sum" is essentially a running total as we go through the terms in a series one by one. The series converge if this running total converge.
A geometric series is a series of the form \[ a + ar + ar^2 + ar^3 + \cdots = \sum_{k=0}^\infty a r^k \] for some real numbers $a \ne 0$ and $|r| \ne 0, 1$.
This series show up in many area of mathematics. One early appearance is in Archimedes' solution to the problem of area bounded by a parabola.
Problem.
Solution. We will start with $n$-th partial sum $S_n$ and $r$ times $S_n$ \begin{align*} S_n &= a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} \\ rS_n &= \hspace{3.6ex} ar + ar^2 + ar^2 + \cdots + ar^{n-1} + ar^n. \end{align*}
The two have many common terms, so \[ S_n - rS_n = a - ar^n = a (1-r^n), \quad\text{i.e.,}\quad (1 - r)S_n = a(1 - r^n). \]
Since we assumed $r \ne 1$. Therefore, \begin{align*} S_n &= \frac{a(1 - r^n)}{1-r}, & &\text{and} & \lim_{n \to \infty} S_n &= \lim_{n \to \infty} \frac{a(1 - r^n)}{1-r} % &= \frac{a(1 - \lim_{n \to \infty}(r^n))}{1-r} = \frac{a}{1-r} \end{align*} if and only if $|r| < 1$. Otherwise the limit does not exist.
The geometric series $a + ar + ar^2 + ar^3 + \cdots$ converges to $\frac{a}{1-r}$ if and only if $|r| < 1$. Otherwise, this series diverges.
Problem. Consider the series $\sum_{k=1}^\infty \frac{1}{k(k+1)}$. Determine its convergence property and evaluate this series if it converges.
Solution. A useful observation is that, via partial fraction decomposition, we can see $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$.
\begin{align*} &\sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) \\ &= \left[ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots+ \left( \frac{1}{n} - \frac{1}{n+1} \right) \right]\\ &= \left[ \frac{1}{1} + \left( - \frac{1}{2} + \frac{1}{2} \right) - \left( \frac{1}{3} + \frac{1}{3} \right) + \cdots+ \left( -\frac{1}{n} + \frac{1}{n} \right) - \frac{1}{n+1} \right]\\ &= \left[ \frac{1}{1} - \frac{1}{n+1} \right] \end{align*}
We have computed that the partial sum is \[ S_n = 1 - \frac{1}{n+1} \] So \[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left[ \frac{1}{1} - \frac{1}{n+1} \right] = 1 - \lim_{n \to \infty} \frac{1}{n+1} = 1 \]
Therefore, we can conclude that the series $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ converges to 1.
Such a series is called a telescoping series because its partial sum collapse due to cancellation of consecutive terms leaving only a few terms.
In general, the idea is to
We come back to the series \[ \sum_{k=0}^\infty (-1)^k = 1 + (-1) +1 + (-1) + \cdots. \] Is it convergent? Is it meaningful to talk about its value?
\[ S_n = \sum_{k=0}^{n-1} (-1)^k = \begin{cases} 0 & \text{if $n$ is even} \\ 1 & \text{if $n$ is odd} \end{cases} \quad\text{so} \lim_{n \to \infty} S_n \text{ does not exist}. \]
Therefore the series $\sum_{k=0}^\infty (-1)^k$ is divergent.
The harmonic series is the series \[ \sum_{k=1}^\infty \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots. \]
Exercise. Use the fact that \begin{align*} &1+\frac{1}{2}+\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} +\frac{1}{7} +\frac{1}{8} + \cdots \\ >\;&1+\frac{1}{2}+\frac{1}{\color{red}4} + \frac{1}{4} + \frac{1}{\color{red}8} + \frac{1}{\color{red}8} +\frac{1}{\color{red}8} +\frac{1}{8} + \cdots \end{align*} to show the harmonic series diverges.
It is useful benchmark case to which other series can be compared.