Sequences and Series

Donovan D Chang

6 Sequences and Series

A sequence is a list of numbers in a definite order:

$a_1,\,a_2,\,a_3,\,\cdots,\,a_n,\,\cdots$

Usually sequences are written inside curly brackets:
$\{a_n\}$ or $\{a_1,\,a_2,\,a_3,\,\cdots,\,a_n,\,\cdots\}$

The limit of a sequence $\{a_n\}$ , $\lim_{n\to\infty}a_n$ , may or may not exist. We say a sequence converges or is convergent when the limit exists, and diverges or is divergent when the limit does not exist. For example, the sequence $a_n=n$ diverges, since $\lim_{n\to\infty}a_n=\infty$ , and the sequence $b_n=\frac{1}{n}$ diverges, since $\lim_{n\to\infty}b_n=0$ .

Some important traits of sequences are defined as follows:

Definition. (Monotonic Sequence)
If $a_n < a_{n+1}$ for all $n\in \mathbb{N}$ , then the sequence $\{a_n\}$ is increasing, and likewise, if $a_n > a_{n+1}$ for all $n\in \mathbb{N}$ , then the sequence $a_n$ is decreasing. Either increasing or decreasing, the sequence is called monotonic.

Definition. (Bounded Sequence)
If $a_n \leq M$ for all $n\in \mathbb{N}$ , then the sequence $\{a_n\}$ is bounded above, and likewise, if $a_n \geq m$ for all $n\in \mathbb{N}$ , then the sequence $\{a_n\}$ is bounded below. Either bounded above or below, $\{a_n\}$ is a bounded sequence.

Consequently, we arrive at a theorem.

Theorem. (Monotonic Sequence Theorem).
If a sequence is bounded and monotonic, then the sequence is convergent.

Proof.
Suppose a sequence is divergent. Then, the sequence has no bound or is alternating – as in a repetition of a certain loop or as in the decimals of an irrational number. Alternating sequence is not monotonic. Therefore, a divergent sequence has no bound or not monotonic. Then, taking the contrapositive, if a sequence is bounded and monotonic, then the sequence is convergent.

It is interesting to see each partial sum up to the $n$ -th term of a sequence $\{a_n\}$ generates another sequence $\{s_n\}$ . We call this $\{s_n\}$ , a series, where $s_n=\sum_{i=1}^{n}a_i$ . For convergent series, we write

$\sum_{n=1}^{\infty}a_n=s$

There are in fact numerous kind of series. Let us begin our exploration with a geometric series.

Geometric series takes the form

$\sum_{n=1}^{\infty}ar^{n-1}$

where $a \neq 0$ .

Let us find a condition for convergence of a geometric series.

$\begin{align*} s_n &= a+ar+ar^2+\cdots +ar^{n-1}\\ rs_n &= \quad\;\;\: ar+ar^2+\cdots +ar^{n-1}+ar^n \end{align*}$

Then, $(1-r)s_n = a\left(1-r^n \right)$ . Note that $\lim_{n\to\infty}r^n=0$ for $r\in(-1,1)$ . Therefore, for $|r|<1$ ,

$s=\frac{a}{1-r}$

Let us utilize the geometric series formula with an example.

Example. Evaluate $\sum_{n=1}^{\infty} \frac{3\cdot 2^{2n}}{9^n}$ .

$\begin{align*} \sum_{n=1}^{\infty} \frac{3\cdot 2^{2n}}{9^n} &= \sum_{n=1}^{\infty} 3\left(\frac{4}{9}\right)^n\\ &= \frac{3\cdot \frac{4}{9}}{1-\frac{4}{9}}\\ &= \frac{12}{5} \end{align*}$

Another type of interesting series is a telescoping series. Its name originates from the cancellation involved in the solution path and only a handful of terms remain as a result, as a telescope is collapsed. Again, let us illustrate with an example.

Example. Find the sum $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ if convergent.

$\begin{align*} &\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\\ = &\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1}\right)\\ = &\lim_{n \to \infty} \left[ \left( 1-\frac{1}{2}\right) + \left( \frac{1}{2} - \frac{1}{3}\right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{n}-\frac{1}{n+1}\right)\right]\\ = &\lim_{n \to \infty} \left( 1-\frac{1}{n+1}\right)\\ = &1 \end{align*}$

Seemingly a similar one, called a harmonic series, in fact diverges to $\infty$ . Let us examine.

Example. Find the sum $\sum_{n=1}^{\infty} \frac{1}{n}= 1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n}+\cdots$ if convergent.

Note that $1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n}+\cdots$ is in fact the right-hand Riemann sum of $y=\frac{1}{x}$ to estimate $\int_{1}^{\infty} \frac{1}{x}\,dx$ .

Therefore,

$\begin{align*} \sum_{n=1}^{\infty} \frac{1}{n} &= 1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n}+\cdots\\ &> \int_{1}^{\infty} \frac{1}{x}\,dx\\ &= \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x}\,dx\\ &= \lim_{t \to \infty} \left[ \ln|x|\right]_1^t\\ &= \lim_{t \to \infty} \ln|t|\\ &= \infty \end{align*}$

Therefore, $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges to $\infty$ .

The technique we just used in determining the convergence of the harmonic series is called an integral test (for the convergence of a series), when a given series can be interpreted as a Riemann sum of a function $f(x)$ and thus compared with $\int_{1}^{\infty} f(x)\,dx$ .

Let us further examine the series in the form of $\sum_{n=1}^{\infty}\frac{1}{n^p}$ by conducting an integral test. What would be the condition for this series to converge?

Let $p\neq 1$ . Then,

$\begin{align*} \sum_{n=1}^{\infty}\frac{1}{n^p} &= 1+\frac{1}{2^p}+\frac{1}{3^p}+\cdots + \frac{1}{n^p}+\cdots\\ &> \int_{1}^{\infty} \frac{1}{x^p}\,dx\\ &= \lim_{t \to \infty} \int_{1}^{t} x^{-p}\,dx\\ &= \lim_{t \to \infty} \left[ \frac{x^{-p+1}}{-p+1}\right]_{1}^{t}\\ &= \lim_{t \to \infty} \frac{1}{1-p}\left[ \frac{1}{t^{p-1}}-1\right] \end{align*}$

1. $p<1$
Then, $\lim_{t \to \infty} \frac{1}{1-p}\left[ \frac{1}{t^{p-1}}-1\right] = \infty$

2. $p>1$
Then, $\lim_{t \to \infty} \frac{1}{1-p}\left[ \frac{1}{t^{p-1}}-1\right] = \frac{1}{p-1}$

Therefore, the condition for the convergence of the $p$ -series $\sum_{n=1}^{\infty}\frac{1}{n^p}$ can be summarized as:

The $p$ -series is convergent if and only if $p>1$ ; otherwise divergent.

Now we come to think what is a condition for any series to converge. Let us suppose a series converges, i.e. there exists $s$ such that $\lim_{n \to \infty}s_n=s$ . Also, by definition, we can write the $n$ -th term of the sequence $\{a_n\}$ as $a_n=s_n-s_{n-1}$ . Then,

$\begin{align*} \lim_{n \to \infty}a_n &= \lim_{n \to \infty}\left(s_n-s_{n-1}\right)\\ &= s-s\\ &= 0 \end{align*}$

Above can be succinctly summarized as follows:

Theorem. If the series $\sum_{n=1}^{\infty}a_n$ is convergent, then $\lim_{n \to \infty}a_n=0$ .

In other words, taking the contrapositive of the theorem,

Corollary. (Test for Divergence)
If $\lim_{n \to \infty}$ does not exist or $\lim_{n \to \infty} \neq 0$ , then the series $\sum_{n=1}^{\infty}a_n$ is divergent.

To examine the convergence of a given series, so far we have conducted integral tests. The comparison test is essentially the same in that we compare a given series with a known series that converges or diverges. For example,

Example. Determine the convergence of $\sum_{n=1}^{\infty}\frac{1}{3^n+1}$ .

In fact, we know $\sum_{n=1}^{\infty}\frac{1}{3^n}$ converges and $\sum_{n=1}^{\infty}\frac{1}{3^n+1}<\sum_{n=1}^{\infty}\frac{1}{3^n+1}$ . Therefore,

$\begin{align*} \sum_{n=1}^{\infty}\frac{1}{3^n+1} &< \sum_{n=1}^{\infty}\frac{1}{3^n+1}\\ &= \frac{\frac{1}{3}}{1-\frac{1}{3}}\\ &= \frac{1}{2} \end{align*}$

Therefore, the series $\sum_{n=1}^{\infty}\frac{1}{3^n+1}$ converges (to a value smaller than $\frac{1}{2}$ ).

The technique we used in fact is called a comparison test. To formally write,

Theorem. (Comparison Test)
For converging series $\sum a_n$ and $\sum b_n$ with positive terms,

If $\sum a_n$ diverges, then $\sum b_n$ diverges.
If $\sum b_n$ converges, then $\sum a_n$ converges.

Now, suppose we are interested in determining the convergence of $\sum_{n=1}^{\infty}\frac{1}{3^n-1}$ . In this case, the comparison test $\sum_{n=1}^{\infty}\frac{1}{3^n-1} > \sum_{n=1}^{\infty}\frac{1}{3^n+1}$ does not help. Instead, we can utilize another technique called the limit comparison test.

Theorem. (Limit Comparison Test)
For converging series $\sum a_n$ and $\sum b_n$ with positive terms, if
$\lim_{n \to \infty} \frac{a_n}{b_n}=c\in \mathbb{R}^+$
then the convergence of both series is the same, i.e. either both diverge or both converge.

Therefore, we conclude $\sum_{n=1}^{\infty}\frac{1}{3^n-1}$ is convergent, since

$\lim_{n \to \infty} \frac{\frac{1}{3^n}}{\frac{1}{3^n-1}} =\lim_{n \to \infty}\left( 1-{\frac{1}{3^n}}\right) =1$

Let us consider a different kind of series, an alternating series, having $(-1)^n$ within its term. A series $\sum_{n=1}^{\infty}\frac{(-1)^n}{3^n-1}$ is an e example of alternating series. In fact, there is a test for convergence for certain types of alternating series.

Theorem. (Alternating Series Test)
For decreasing sequence $b_n$ with $\lim_{n \to \infty}b_n=0$ , its alternating series
$\sum_{n=1}^{\infty}(-1)^{n-1}b_n$
is convergent.

Let us illustrate with an example.

Example. Determine the convergence of $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$ .

Let $b_n=\frac{1}{n}$ . Then,

1. $b_n$ is a decreasing sequence, since $\frac{1}{n+1}<\frac{1}{n}$ for all $n\in\mathbb{N}$
2. $\lim_{n \to \infty}b_n=\lim_{n \to \infty}\frac{1}{n}=0$

Therefore, by alternating series test, the series $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$ is convergent.

Now, let us introduce the concept of absolute convergence.

Definition. (Absolute Convergence)
A series $\sum a_n$ is absolutely convergent if $\sum |a_n|$ is convergent.

The notion of absolute convergence was needed for the ratio test. The ratio test is highly useful a test in determining the convergence of a series. The ratio is defined as follows:

Theorem. (Ratio Test)
Given
$\lim_{n \to \infty}\left\vert \frac{a_{n+1}}{a_n} \right\vert=L$

If $L<1$ , then the series $\sum_{n=1}^{\infty} a_n$ is absolutely convergent.
If $L>1$ , then the series $\sum_{n=1}^{\infty} a_n$ is divergent.
If $L=1$ , then the test is inconclusive.

Let us illustrate with an example.

Example. Determine the convergence of $\sum_{n=1}^{\infty}\frac{n^n}{n!}$ .

Let us conduct the ratio test. Then,

$\begin{align*} \lim_{n \to \infty}\left\vert \frac{\frac{(n+1)^{n+1}}{(n+1)!}}{\frac{n^n}{n!}} \right\vert &= \lim_{n \to \infty} \left( \frac{n+1}{n}\right)^n\\ &= \lim_{n \to \infty} \left( 1+\frac{1}{n}\right)^n\\ &= e\\ &>1 \end{align*}$

Therefore, by the ratio test, the series $\sum_{n=1}^{\infty}\frac{n^n}{n!}$ is divergent.

We conclude this section introducing a similar, but another useful test, the root test.

Theorem. (Root Test)
Given
$\lim_{n \to \infty} \sqrt[n]{|a_n|}=L$

If $L<1$ , then the series $\sum_{n=1}^{\infty} a_n$ is absolutely convergent.
If $L>1$ , then the series $\sum_{n=1}^{\infty} a_n$ is divergent.
If $L=1$ , then the test is inconclusive.

6 Sequences and Series

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