MAT187 L09 - Sequences and Series

An infinite sequence is an ordered list of terms of form:

a_{1}, a_{2}, a_{3}, \dots, a_{n}, \dots

On the other hand, an infinite series is a sum of terms in a sequence, in form

\sum_{i = 0}^{n} a_{i} = a_{0} + a_{1} + a_{2} + \dots + a_{n}

Warning! Do not mix up sequences and series!

Sequences are lists, series are sums of sequences.

The limit of a series as $i$ approaches infinity can either converge or diverge.

\begin{aligned} lim_{n \to \infty} (\sum_{i = 0}^{n} a_{i}) & = c, c ϵ R ...[series converges] \\ lim_{n \to \infty} (\sum_{i = 0}^{n} a_{i}) & \to \pm \infty ...[series diverges] \\ lim_{n \to \infty} (\sum_{i = 0}^{n} a_{i}) & = D N E ...[series diverges] \end{aligned}

The last case could be something like:

lim_{n \to \infty} (\sum_{i = 0}^{n} (- 1)^{i})

...which doesn't exist because it oscillates.

Comparison Test

A way of finding the behaviour of an infinite series by comparing it to a simpler series that preserves certain properties of the original sum.

In other words, if a function $P (x)$ converges and is greater than the function of interest, $f (x)$ , as $x$ approaches infinity, then $f (x)$ converges as well.
Similarly, if $P (x)$ diverges and is less than $f (x)$ as $x$ approaches infinity, then $f (x)$ diverges as well.
Only works if both series have positive terms.

You usually determine this by taking a difficult function and simplifying it, then trying to see if you can manipulate inequalities to prove that the simpler function is either greater than or less than the original function as $x$ approaches infinity.

Example
Consider $\sum_{n = 0}^{\infty} \frac{n - 1}{n^{2} + 3 n}$ . Is it a converging or diverging infinite series?

A simpler function to examine would be $\frac{1}{n}$ .
$n - 1 \geq \frac{1}{2} n$ as $n \to \infty$ , and
$n^{2} + 3 n \leq 2 n^{2}$
Since the numerator is always greater, and the denominator is always smaller, then:
$\sum_{n = 0}^{\infty} \frac{n - 1}{n^{2} + 3 n} \geq \frac{\frac{1}{2} n}{2 n^{2}} = \frac{1}{n}$
We know properties of $\frac{1}{n}$ . Since it diverges, and $\sum_{n = 0}^{\infty} \frac{n - 1}{n^{2} + 3 n} \geq \sum_{n = 0}^{\infty} \frac{1}{n}$ , then $\sum_{n = 0}^{\infty} \frac{n - 1}{n^{2} + 3 n}$ diverges as well.

Limit Comparison Test

Given $\sum a_{n}$ , $\sum b_{n}$ series with positive terms, let $L = lim_{k \to \infty} \frac{a_{k}}{b_{k}}$ .

If $0 < L < \infty$ , then $\sum a_{n}$ and $\sum b_{n}$ both diverge, or both converge.
If $L = 0$ , and $\sum b_{n}$ converges, so too does $\sum a_{n}$ .
- i.e. $\sum b_{n}$ "overpowers" $\sum a_{n}$ .
If $L \to \infty$ and $\sum b_{n}$ diverges, so too does $\sum a_{n}$ .
- i.e. $\sum a_{n}$ "overpowers" $\sum b_{n}$ .

Example
Consider $\sum_{n = 0}^{\infty} \frac{n - 1}{n^{2} + 3 n}$ . Is it a converging or diverging infinite series?

We know that $\frac{1}{n}$ is a diverging series. Thus, we can use limit comparison test.
Let $L = lim_{k \to \infty} \frac{\frac{k - 1}{k^{2} + 3 k}}{\frac{1}{k}}$
$= lim_{k \to \infty} (\frac{k (k - 1)}{k (k + 3)})$
$= lim_{k \to \infty} \frac{k - 1}{k + 3}$
$= lim_{k \to \infty} \frac{- 4 + (k + 3)}{k + 3}$
$= lim_{k \to \infty} \frac{- 4}{k + 3} + lim_{k \to \infty} \frac{k + 3}{k + 3}$
$= 1$
Therefore, by limit comparison test, we know that $\sum_{n = 0}^{\infty} \frac{n - 1}{n^{2} + 3 n}$ follows the same divergence as $\frac{1}{n}$ .

Ratio Test

If $\sum a_{k}$ is a series with positive terms, let $L = lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}}$ .

If $0 \leq r < 1$ , then $\sum a_{k}$ converges.
If $r > 1$ , then $\sum a_{k}$ diverges.
If $r = 1$ or $lim_{n \to \infty}$ does not exist, test is inconclusive.

Note: I messed up and this is not content that is important for Midterm 1
Oops! To see the actual note for ratio test: MAT187 L14-15 - Power Series Convergence