Series

let ( $b_{n}$ ) be a sequence. An infinite series is:

\sum_{n = 1}^{\infty} b_{n}

where $s_{m} = \sum_{n = 1}^{m} b_{n}$ is partial sum.

Convergence of a Series

A series converges if its partial sums converge

\sum_{n = 1}^{\infty} \frac{1}{n^{2}}

\begin{array}{lcl} s_{m} & = & 1 + \frac{1}{2 \cdot 2} + \frac{1}{3 \cdot 3} + \frac{1}{4 \cdot 4} + \cdot \cdot \cdot + \frac{1}{m^{2}} \\ < & 1 + \frac{1}{2 \cdot 1} + \frac{1}{3 \cdot 2} + \frac{1}{4 \cdot 3} + \cdot \cdot \cdot + \frac{1}{m (m - 1)} \\ = & 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \cdot \cdot \cdot + (\frac{1}{(m - 1)} - \frac{1}{m}) \\ = & 1 + 1 - \frac{1}{m} \\ < & 2. \end{array}

By monotone convergence theorem the sequence ( $s_{m}$ ) is bounded and strictly increasing so it converges.

\sum_{n = 1}^{\infty} \frac{1}{n}

\begin{array}{rcl} s_{2^{k}} & = & 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \dots + \frac{1}{8}) + \dots + (\frac{1}{2^{k - 1} + 1} + \dots + \frac{1}{2^{k}}) \\ > & 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \dots + \frac{1}{8}) + \dots + (\frac{1}{2^{k}} + \dots + \frac{1}{2^{k}}) \\ = & 1 + \frac{1}{2} + 2 (\frac{1}{4}) + 4 (\frac{1}{8}) + \dots + 2^{k - 1} (\frac{1}{2^{k}}) \\ = & 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots + \frac{1}{2} \\ = & 1 + k (\frac{1}{2}), \end{array}

So ( $s_{2^{k}}$ ) is unbounded so it does not converge.

Let $x_{n} = r^{n}$ $r \in R$

\begin{array}{r} \sum_{n = 0}^{\infty} r^{n} = 1 + r + r^{3} + \dots \end{array}

If $r = 1$ then $lim S_{n} = lim \underset{n times}{(} 1 + 1 + 1 + \dots + 1)$
Which diverges
If $r \neq 0$

\begin{aligned} (1 - r) (1 + r + r^{2} \dots + r^{n}) & = 1 - r^{n + 1} \\ S_{n} & = \frac{1 - r^{n + 1}}{1 - r} \end{aligned}

If $| r | < 1 ⟹ lim r^{n} = 0$
If $| r | > 1 ⟹ lim r^{n}$ diverges

If $| r | < 1 ⟹$

\begin{aligned} lim S_{n} & = lim \frac{1 - r^{n + 1}}{1 - r} \\ = \frac{1}{1 - r} \end{aligned}

Using the algebraic limit theorems of sequences we get

\sum_{k = 0}^{\infty} a r^{k} = \frac{a}{1 - r}

iff $| r | < 1$

$\sum x_{n}$ converges if and only if for every $ε > 0$ there exists a $K \in N$ so that if $m > n \geq K$ then

| S_{n} + S_{m} | < ε

If $\sum_{n = 1}^{\infty} | a_{n} |$ converges then $\sum_{n = 1}^{\infty} a_{n}$ converges

$\sum a_{n}$ converges but $\sum | a_{n} |$ does not
Can't re

Suppose ( $b_{n}$ ) is decreasing and $b_{n} \geq 0$ for all $n \in N$ then $\sum_{n = 1}^{\infty} b_{n}$ converges if and only if

\sum_{n = 0}^{\infty} 2^{n} b_{2^{n}} = b_{1} + 2 b_{2} + 4 b_{4} + 8 b_{8} + 16 b_{16} + \cdot \cdot \cdot

converges.

If $\sum_{n = 1}^{\infty} 2^{n} b_{2^{n}}$ converges then the partial sums are bounded by $M > 0$ .

Since $b_{n} \geq 0$ we know its partial sums $s_{m}$ are increasing. Just need to show that its bounded.

$m \leq 2^{k + 1} - 1$ . Then $s_{m} \leq s_{2^{k + 1} - 1}$

\begin{aligned} s_{2^{k + 1} - 1} & = b_{1} + (b_{2} + b_{3}) + (b_{4} + b_{5} + b_{6} + b_{7}) + \dots + (b_{2^{k}} + \dots + b_{2^{k + 1} - 1}) \\ \leq b_{1} + (b_{2} + b_{2}) + (b_{4} + b_{4} + b_{4} + b_{4}) + \dots + (b_{2^{k}} + \dots + b_{2^{k}}) \\ = b_{1} + 2 b_{2} + 4 b_{4} + \dots + 2^{k} b_{2^{k}} = t_{k} \end{aligned}

$s_{m} \leq t_{k} \leq M$