Series

Let $(u_n)$ be a sequence, and a series (a new sequence) can be defined from it such that: $$

$$s_n=\sum_{k=1}^{n}{u_k}$$

Convergence

If $(s_n)$ is converging: $$

$$\lim_{n\to \infty}{s_n} = \lim_{n\to \infty}{\sum_{k=1}^n {u_k}} = \sum_{k=1}^\infty u_k = S\in\mathbb{R}$$

Absolutely Converging

$\sum_{k=1}^{n}{u_k}$ is absolutely converging iff $\sum_{k=1}^{n}{\lvert u_k \rvert}\text{ is converging}$.

$$\sum_{k=1}^{n}{\lvert u_k \rvert}\text{ is converging} \implies \sum_{k=1}^{n}{u_k}\text{ is converging}$$

Proof Hint

Use this inequality:

$$0 \le \;\lvert u_k \rvert - u_k\; \le 2 \lvert u_k \rvert$$

Theorem

A series $s_n$ is absolutely converging to $s$ iff rearranged series of $s_n$ converges to $s$.

Conditionally Converging

$\sum_{k=1}^{n}{u_k}$ is condtionally converging iff: $$

$$\sum_{k=1}^{n}{\lvert u_k \rvert}\text{ is diverging}\;\;\;\text{and}\;\sum_{k=1}^{n}{u_k}\text{ is converging}$$

Theorem

Suppose $s_n$ is a conditionally converging series. Then: $$

1. Sum of all the positive terms limits to $\infty$
2. Sum of all the negative terms limits to $-\infty$
3. $s_n$ can be rearranged to have the sum:
   • Any real number $x$
   • $\infty$
   • $-\infty$
   • Does not exist

Terms limit to 0

$$\sum_{k=1}^{n}{u_k}\text{ is converging} \implies \lim_{k\to\infty}{u_k} = 0$$

The converse is known as the divergence test:

Grouping

Suppose $\sum u_k$ is a given series. If $v_n$ is formed by grouping a finite number of adjacent terms $u_k$, then $\sum v_k$ is a grouping of the given series.

Rearrangement

Suppose $\sum u_k$ is a given series. If there is a bijection sequence $k_n$ such that $v_n = u_{k_n}$, then $\sum v_n$ is a rearrangement of the given series.