There are a few minor errors in this section, but I am too careless to fix them because this content is very boring.

Series

Given a sequence $\{a_n\}$ in $\mathbb{C}$, consider the partial sums $s_n = \sum_{k=1} ^n a_k$. The sequence formed by these partial sums is denoted $\Sigma a_n$ and is called the infinite series associated with $\{a_n\}$.

$$ \{a_n\} \rightarrow \{s_n\} \equiv \Sigma a_n \text{ with } s_n = \sum _{k=1}^n a_k $$

Moreover, we say a series converges if there exists a limit $s$ for which $\lim_{n \rightarrow \infty} \Sigma a_n = s$, for which $s$is denoted $s = \sum_{k=1} ^{\infty} a_n$.

Convergent Series have Vanishing Terms. If a series $\Sigma a_n$ converges, then $\{a_n\}$ converges to $0$.

$$ \Sigma a_n \text{ converges} \implies \lim _{n\rightarrow \infty} a_n = 0 $$

Proof </aside>

Convergent Series iff Weighted Power of 2 Subseries Converges.

$$ \Sigma a_n \text{ converges} \iff \Sigma a_{2^n} 2^n \text{ converges} $$

</aside>

Bounded Partial Sums with Decreasing, Vanishing, and Nonnegative Coefficients Implies Convergence. If $\Sigma a_n$ is a bounded sequence and $\{ b_n\}$ is monotonically decreasing with $b_n \geq 0$ for all $n$ and $b_n \rightarrow 0$, then $\Sigma a_n b_n$ converges.

$$ \begin {cases} \Sigma a_n \text{ bounded } \\ b_0 \geq b_1 \geq \dots \geq 0 \\ \lim _{n\rightarrow \infty } b_n = 0 \end{cases} \implies \Sigma a_n b_n \text{ converges} $$

As a corollary to this, the series $\Sigma \frac{(-1)^n}{n}$ converges.

</aside>

Geometric Series. Given $x \in \R$, the series formed by the sequence $\{x^n\}$ converges or diverges based on the following criterion:

$$ \begin{cases} 0\leq x < 1 \implies \Sigma x^n \rightarrow \frac{1}{1-x} \\ x>1 \implies \Sigma x^n \text{ diverges }\end{cases} $$

Proof </aside>

$p$-Series. Given $p \in \R$, the series formed by the sequence $\{\frac{1}{n^p}\}$ converges or diverges based on the following criterion:

$$ \begin{cases} p > 1 \implies \Sigma \frac{1}{n^p} \text{ converges} \\ p \leq 1 \implies \Sigma \frac{1}{n^p} \text{ diverges}\end{cases} $$

Proof </aside>

Comparison Test. Given a series $\Sigma a_n$, if there is a bigger series that converges, then $\Sigma a_n$ converges and if there is a smaller series that diverges, then $\Sigma a_n$ diverges.

$$ \forall n \geq N, |a_n| \leq c_n \implies [\Sigma c_n \text{ converges } \implies \Sigma a_n \text{ converges}] $$

$$ \forall n \geq N, |a_n| \geq c_n \geq 0 \implies [\Sigma c_n \text{ diverges } \implies \Sigma a_n \text{ diverges }] $$

Proof </aside>

Root Test. Given a series ****$\Sigma a_n$, define $\alpha = \limsup _{n\rightarrow \infty} |a_n|^{\frac{1}{n}}$.

$$ \begin{cases} \alpha < 1 \implies \Sigma a_n \text{ converges } \\ \alpha > 1 \implies \Sigma a_n \text{ diverges } \\ \alpha = 1 \implies \text{no information}\end{cases} $$

Proof </aside>

Ratio Test. Given a series $\Sigma a_n$ consider a sequence $\{|\frac{a_{n+1}}{a_n}|\}$.

$$ \begin{cases} \limsup {n\rightarrow \infty } |\frac{a{n+1}}{a_n}| < 1 \implies \Sigma a_n \text{ converges} \\ \exists N \in \N : \forall n \geq N, |\frac{a_{n+1}}{a_n}| \geq 1 \implies \Sigma a_n \text{ diverges}\end{cases} $$

</aside>

Power Series

Given a sequence $\{a_n\}$ in $\mathbb{C}$ and $z \in \mathbb{C}$, the associated power series is a series of the form:

$$ \sum _{n=0} ^{\infty} a_n z^n $$

Radius of Convergence. The radius of convergence $R = \frac{1}{\limsup _{n\rightarrow \infty} |a_n|^{\frac{1}{n}}}$ of a power series $\Sigma a_n z^n$ governs the set of $z$ which permits convergence.

$$ \begin{cases}|z| < R \implies \Sigma a_n z^n \text{ converges} \\ |z| > R \implies \Sigma a_n z^n \text{ diverges} \\ |z| = R \implies \text{no information}\end{cases} $$

Proof </aside>

Series

Power Series

Absolute Convergence