# Wednesday, March 27, 2013 Introduction to series

Proof: (ii) Let $y \in \mathbb R$ be fixed and consider $G(x) = E(x + y) / E(y)$. We will show that $G(x) = E(x)$. Observe that

$$G'(x) = \frac{E'(x+y)}{E(y)} = \frac{E()}{}$$

$= G(x)$ and $G(0) = E(y) / E(y) = 1$ so $G(x) = E(x)$

(iii) by induction on (ii), $E(nx) = E(x) + E(x) + \ldots + E(x) = (E(x))^n$ for $x \in \mathbb N$. Let $x = \frac{1}{n}$

And then it was all erased.

## 1Series¶

A sequence is a sum of terms $\sum a_n$ in a sequence $(a_n)$.

Definition: Given a series $\displaystyle \sum_{n=1}^\infty a_n$ we define the sequence of partial sums $(S_n)$ such that $S_1 = a_1$, $S_2 = a_1 + a_2$, etc. If $(S_n)$ converges we say the series converges and

$$\sum_{n=1}^\infty a_n = \lim S_n$$

Example: $\displaystyle \sum_{n=1}^\infty a^n = 1 + a + a^2 + \ldots$. $S_1 = 1$, $S_2 = 1 + a$, $S_3 = 1 + a + a^2$, and so on, so $S_n = 1 + a + a^2 + \ldots + a^{n-1}$.

$$S_n = (1 + a + a^2 + \ldots + a^{n-1})\left(\frac{1-a}{1-a}\right) = \frac{1 + a + a^2 + \ldots + a^{n-1} - a - a^2 - \ldots - a^{n-1} - a^n}{1-a}$$

This is a telescoping sum, as almost all the terms in the numerator cancel giving us $\frac{1-a^n}{1-a}$.

Now if $|a| < 1$, $|a|^n \to 0$ thus for a given $\epsilon > 0$ there exists some $N \in \mathbb N$ such that if $n \geq N$, $|a|^n < \epsilon|1-a|$ and so

$$\left|S_n - \frac{1}{1-a}\right| = \left| \frac{1-a^n}{1-a} - \frac{1}{1-a}\right| = \left|\frac{-a^n}{1-a}\right| = \frac{|a|^n}{|a-a|} < \epsilon\frac{|1-a|}{|1-a|} = \epsilon$$

i.e. $S_n \to \frac{1}{a-1}$ if $|a| < 1$ show that if $|a| \geq 1$ then either $S_n \to \infty$ or $S_n$ does not converge.

Definition: Series of the form $\displaystyle \sum_{n=0}^\infty a^n$ are known as geometric series.

Theorem: $\displaystyle \sum_{n=0}^\infty a^n$ converges to $\frac{1}{1-a}$ if $|a|<1$ and diverges otherwise. If $|a| < 1$ then $\displaystyle \sum_{n=k}^\infty a^n = \frac{a^k}{1-a}$.

Example: $\displaystyle \sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right)$.

$S_1 = 1 - \frac{1}{2}$, $S_2 = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = 1 - \frac{1}{3}$, and etc so $S_n = 1 = \frac{1}{n+1}$. This can be shown by induction. Thus $S_n \to 1$ and so $\displaystyle \sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$.

Theorem: If $\displaystyle \sum_{n=1}^\infty a_n$ converges then $a_n \to 0$

Proof: Let $S_n$ be the sequence of partial sums. By assumption $S_n \to L$. It's clear that $S_n - S_{n-1} = a_n$. As $n \to \infty$, $S_n \to L$ and $S_{n-1} \to L$ thus $a_n = S_n - S_{n-1} \to L - L = 0$.

Cauchy Criterion: $\displaystyle \sum a_n$ converges $\iff$ for $\epsilon > 0$ given, there exists some $N \in \mathbb N$ such that if $m > n \geq N$ then $|S_m - S_n| = |a_{n+1} + a_{n+2} + \ldots + a_m| < \epsilon$

Example: $\displaystyle \sum_{n=1}^\infty \frac{1}{n}$, or the harmonic series, diverges. We can show this by first looking at the sequence of partial sums. $S_1 = 1$. $S_2 = 1 + \frac{1}{2}$. $S_4 = S_{2^2} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 2$. Extending this, $S_{2^n} \geq 1 + \frac{n}{2} \to \infty$.