Sequences and series

Sequences

A sequence is an infinite list of numbers defined by it's $n$th term.
A sequence converges to a limit if its terms eventually get close to a fixed number.

Convergent sequences

A sequence ($a_n$) of real numbers is said to converge to a limit $l \in \mathbb{R}$ if for every $\epsilon > 0$ there is an integer $N$ whith $|a_n - l| < \epsilon$ for all $n > N$.

Combination rules for convergent sequences

Bounds

A sequence ($a_n$) is said to be bounded above if there is a number $U$ such that $a_n \leq U$ for all $n$.
A sequence is bounded below if there is an $L$ such that $L \leq a_n$ for all $n$.

Basic properties of convergent sequences

1. A convergent sequence has a unique limit

2. If $a_n \rightarrow l$ then every subsequence of $a_n$ also converges to $l$

3. If $a_n \rightarrow l$ then $|a_n| \rightarrow |l|$

4. The squeeze rule - If $a_n \rightarrow l$ and $b_n \rightarrow l$ and $a_n \leq c_n \leq b_n$ for all $n$ then $c_n \rightarrow l$

5. A convergent sequence is bounded

6. Any increasing sequence bounded above converges

7. Any decreasing sequence bounded below converges

Sequences continued

$r^n$ converges for $0 \leq r < 1$ because it is a decreasing sequence and bounded below by 0, so by 16.4 property 7 it must converge. Therefore $r^n \rightarrow l$. Using some stuff in the slides given $l = 0$, so $r^n \rightarrow 0$.

Divergent sequences

A sequence is said to diverge to infinity if for every $K \in \mathbb{R}$ there is an $N$ with $a_n > K$ whenever $n>N$. This is written as $a_n \rightarrow \infty$.
A non-convergent sequence which does not diverge to $\pm \infty$ is said to oscillate. The combination rules do not apply to sequences which diverge to $\pm \infty$.

Basic convergent sequences

$$ \begin{aligned} &\lim_{n \rightarrow \infty}{\frac{1}{n^p}} = 0 \text{ for any } p>0\\ &\lim_{n \rightarrow \infty}{c^n} = 0 \text{ for any } c \text{ with } |c| < 1\\ &\lim_{n \rightarrow \infty}{c^{\frac{1}{n}}} = 1 \text{ for any } c>0\\ &\lim_{n \rightarrow \infty}{n^pc^n}=0 \text{ for } p>0 \text{ and } |c| < 1\\ &\lim_{n \rightarrow \infty}{\frac{c^n}{n!}} = 0 \text{ for any } c \in \mathbb{R}\\ &\lim_{n \rightarrow \infty}{\left(1+\frac{c}{n}\right)^n} = e^c \text{ for any } c \in \mathbb{R}\end{aligned} $$

Big O notation

If $a_n$ and $b_n$ are sequences of real numbers then $a_n$ is $O(b_n)$ if there are constants $C$ and $N$ with $|a_n| \leq C|b_n|$ for all $n \geq N$.

Recurrences

A recurrence is a rule which defines each term of a sequence using the preceding terms, such as the Fibonacci sequence.

Series

A series $\Sigma a_n$ is a pair of sequences consisting of a sequence of terms $a_n$ and a sequence of partial sums $s_n$.

A series $\Sigma a_n$ converges to the sum $s$ if the sequence $s_n$ of partial sums converges to $s$. This is written as $\Sigma_{n=0}^\infty a_n = s$. A series diverges if it does not converge.

The geometric series $\Sigma r^n$ converges to $\frac{1}{1-r}$ provided $|r|<1$.
Proof:
Let $s_n = 1 + r + r^2 ... + r^n$
Then $rs_n = r(1+r+r^2+...+r^n)$
$s_n-rs_n = 1-r^{n+1}$
$(1-r)s_n = 1-r^{n+1} \rightarrow s_n = \frac{1-r^{n+1}}{1-r}$
$r^{n+1} \rightarrow 0$ when $|r| < 1$ so $s_n = \frac{1}{1-r}$ for $|r|<1$.

The harmonic series $\Sigma_{n=1}^\infty \frac{1}{n}$ diverges. Proof:
Consider the even subsequences of partial sums:
$s_2 = 1 + \frac{1}{2}$
$s_4 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} > 1 + \frac{1}{2} + \frac{1}{2}$
$s_8 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{8} > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$ So in general $s_{2n} > 1+\frac{n}{2}$. This subsequence is unbounded so does not converge. If the subsequence does not converge then the sequence does not converge.

Basic properties of convergent series

1. Sum rule - If $\Sigma a_n$ converges to $s$ and $\Sigma b_n$ converges to $t$ then $\Sigma(a_n+b_n)$ converges to $s+t$.

2. Multiple rule - If $\Sigma a$ converges to $s$ and $\lambda \in \mathbb{R}$ then $\Sigma \lambda a_n$ converges to $\lambda s$.

3. The sequence of a convergent series rule - If series $\Sigma a$ converges then the sequence $a_n$ converges to 0. (This does not work the other way round)

4. The modulus rule - If series $\Sigma |a_n|$ converges then series $\Sigma a_n$ also converges.

Series continued

Comparison test

Suppose that $0 \leq a_n \leq b_n$ for every $n$. If $\Sigma b_n$ converges then so does $\Sigma a_n$. If $\Sigma a_n$ diverges then so does $\Sigma b_n$.

Ratio test

If $\left|\frac{a_{n+1}}{a_n}\right| \rightarrow L$ then if $0 \leq L < 1$ then $\Sigma a_n$ converges, if $L>1$ then $\Sigma a_n$ diverges and if $L = 1$ then the test is inconclusive.

Basic convergent series

1. $\Sigma_{n=0}^\infty r^n = \frac{1}{1-r}$ converges for any $r$ with $|r|<1$

2. $\Sigma \frac{1}{n^k}$ converges for any $k>1$

3. $\Sigma n^kr^n$ converges for $k > 0$ and $|r|<1$

4. $\Sigma_{n=0}^\infty \frac{c^n}{n!} = e^c$ for any $c \in \mathbb{R}$

Basic divergent series

1. $\Sigma \frac{1}{n^k}$ diverges for any $k \leq 1$

Power series

A power series is one of the form $\Sigma a_nx^n$.

If $\Sigma a_nR^n$ converges for some $R \geq 0$ then $\Sigma a_nx^n$ converges for every $x$ with $|x|<R$.

The radius of convergence of a power series $\Sigma a_nx^n$ is the number $R \geq 0$ such that the series converges whenever $|x|<R$ and diverges whenever $|x|>R$.

Properties of power series

Let $f(x) = \Sigma_{n=0}^\infty a_nx^n$ and $g(y) = \Sigma_{n=0}^\infty b_ny^n$ for $x \in (-R_1, R_1)$ and $y \in (-R_2, R_2)$ where $R_1, R_2 > 0$.
If $R = \operatorname{min}(R_1, R_2)$ then if $f(x) = g(x)$ for all $x \in (-R, R)$ then $a_n = b_n$ for each $n$ (equality rule).
Also, for any $x \in (-R, R)$: $f(x) + g(x) = \Sigma_{n=0}^\infty (a_n+b_n)x^n$ (sum rule), $\lambda f(x) = \Sigma_{n=0}^\infty \lambda a_nx^n$ for any $\lambda \in \mathbb{R}$ (multiple rule) and $f(x)g(x) = \Sigma_{n=0}^\infty (a_0b_n + a_1b_{n-1}+...+a_nb_0$ (product rule).

General binomial theorem

For any rational number $q$ and $x \in (-1, 1)$ then $(1+x)^q = \Sigma_{n=0}^\infty \begin{pmatrix} q \ n \end{pmatrix} x^n$ where $\begin{pmatrix} q \ n \end{pmatrix} = \frac{q(q-1)...(q-(n-1)}{n!}$

Decimal representation of real numbers

$\Sigma_{n \geq 0} r^n = \frac{1}{1-r}$. If $a_1, a_2, ...$ is a sequence of decimal digits then $.a_1a_2a_3...$ denotes the real number $\Sigma \frac{a_n}{10^n}$.
$0 \leq \frac{a_n}{10^n} \leq \frac{9}{10^n}$ for each $n$.
$\Sigma \frac{1}{10^n}$ converges when $r > 0.1$ so $\Sigma \frac{9}{10^n}$ converges, so $\Sigma \frac{a_n}{10^n}$ converges by the comparison test.

A terminating decimal has a finite number of decimal places. A non-terminating or repeating decimal has infinite. Both represent a rational number.

To convert the repeating decimal 0.59102102102... as the quotient of two integers, first split it into non-repeating and repeating parts: $\frac{59}{100} + \frac{102}{10^5} + \frac{102}{10^8} ...$. The second part can be rewritten as a geometric series: $\frac{59}{100} + \frac{102}{10^5}\left(1+\frac{1}{10^3} + \frac{1}{10^6}+...\right)$ which converges to $\frac{59}{100} + \frac{102}{10^5}\left(\frac{1}{1-\frac{1}{10^3}}\right)$ which can be simplified to give the quotient of two integers.

A non-terminating decimal representation can also be written as a terminating decimal.
$0.499999999... = 0.5$. $0.499999... = \frac{4}{10} + \frac{9}{10^2} \left(1 + \frac{1}{10}+...\right) = \frac{4}{10}+\frac{9}{10^2} \times \frac{1}{1-\frac{1}{10}} = \frac{4}{10} + \frac{9}{90} = \frac{4}{10} + \frac{1}{10} = \frac{5}{10} = 0.5$.

Irrational numbers are represented by decimal expansions which do not terminate or repeat. If $x$ is not an integer then it lies between two consecutive integer $a_0$ and $a_0+1$. This interval can be divided into 10 equal parts, giving $a_0 + \frac{a_1}{10} < x < a_0 + \frac{a_1+1}{10}$ where $a_1 \in \{0,1,...,9\}$. This new interval can be divided into 10 more intervals where $a_0 + \frac{a_1}{10} + \frac{a_2}{10^2}+...<x<a_0 + \frac{a_1+1}{10} + \frac{a_2+1}{10^2}+...$. Continuing this gives two geometric sequences, one increasing from below and the other increasing from above.

Any rational number can be written in the form $\frac{r}{s} = \frac{q_1}{10} + \frac{q_2}{10^2}+...+\frac{q_n}{10^n} + \frac{1}{10^n} \times \frac{r_n}{s}$ where $q_i$ is one of the digits 0-9.

$10r_i = sq_{i+1}+r_{i+1}$ where $0 \leq r_1 < s$.

Consider $\frac{5}{14}$. $r_0 = 5$ and $s = 14$. This gives $10(5) = 14 \times 3 + 8$ because $50 \div 14 = 3$ remainder $8$. Then $80 = 14 \times 5 + 10$ because $r_1$ was 8 and $80 \div 14 = 5$ remainder 10. This continues until $r_n$ has already appeared in the sequence. This repeated part is the recurring decimal.