# Series

series can be described as the sum of some set of terms of a sequence. The $$n$$th partial sum of a sequence is usually called $$S_n$$. If the sequence being summed is $$s_n$$ we can use sigma notation to define the series:

$S_n=\sum_{i=1}^ns_i$

which just says to sum up the first $$n$$ terms of the sequence $$s$$. If we sum up an infinite number of terms of a sequence we get an infinite series.

## Sequence

sequence ($$s$$) is an enumerated collection of objects in a specified order. Sequences are used in infinitesimal calculus, which largely uses the concept of limit of a sequence. They play a fundamental role in defining the set of real numbers (and in calculus in general), as they represent a basis for the study of functions in the real field.

Like a set, it contains members (also called elements, or terms). The number of elements (possibly infinite) is called the length of the sequence. Unlike a set, the same elements can appear multiple times at different positions in a sequence, and order does matter. Formally, a sequence can be defined as a function whose domain is either the set of the natural numbers (for infinite sequences) or the set of the first $$n$$ natural numbers (for a sequence of finite length $$n$$). The formal writing of the sequences is varied, and changes according to whether they are considered in general discourses, attributable to an axiomatic approach, or specific or calculable sequences are taken into consideration. Since the terms of a sequence are infinite, they cannot all be written explicitly and therefore various expedients are used. For example, to represent a sequence, we often just write some initial terms followed by ellipsis:

$s=(a_{1},a_{2},a_{3},\cdots ,a_{n},\cdots)$

## Arithmetic progression

An arithmetic progression (or arithmetic sequence or arithmetic series) is a sequence of numbers such that the difference between the consecutive terms is constant.

$\sum_{n=0}^{+\infty}a+nd$

that is, if its associated sequence is: $$a_n=a+nd$$ where $$d$$ is called reason, and $$a$$ is the starting point. The reduced sum of the arithmetic series has two equivalent forms (implicit or explicit):

Implicit form: $$\displaystyle{\sum_{i=0}^{n}}a+id=\dfrac{1}{2}(a_n+a_0)(n+1)$$

Explicit form: $$\displaystyle{\sum_{i=0}^{n}}a+id=(n+1)\left(a+\dfrac{1}{2}nd\right)$$

To study the convergence of the arithmetic series we write its reduced $$n$$th in explicit form:

$s_n=\sum_{i=0}^na+id=(n+1)\left(a+\dfrac{1}{2}nd\right)$

$\lim_{n\rightarrow +\infty}s_n=\lim_{n\rightarrow +\infty}(n+1)\left(a+\dfrac{1}{2}nd\right)=\lim_{n\rightarrow +\infty}dn^2$

$\lim_{n\rightarrow +\infty}dn^2=\left\{\begin{matrix} +\infty & if:\;d>0\\ -\infty & if:\;d<0 \end{matrix}\right.$

In any case we have that the arithmetic series diverges.

## Harmonic series

The harmonic series is the divergent infinite series:

$\sum_{n=1}^{+\infty}\dfrac{1}{n^p}$

The harmonic series has two possible characters:

1. convergent if $$p>1$$
2. divergent if $$p\leq 1$$

## Geometric series

A series is called geometric if it occurs in the following form:

$\sum_{n=0}^{+\infty}aq^n$

where $$q$$ is called reason and $$a$$ is the starting point. A convenient way to recognize it immediately is to make the ratio of two consecutive series terms:

$\dfrac{a_{n+1}}{a_n}=\dfrac{aq^{n+1}}{aq^n}=q$

that is, it is constantly the same as reason. Demonstration of the reduced $$n$$th of the geometric series $$a_n=aq^n$$ where $$q\neq 1$$:

$\forall n\in \mathbb{N}\left(\sum_{i=0}^naq^i=a\dfrac{1-q^{n+1}}{1-q}\right)$

$S_0=a\dfrac{1-q^{0+1}}{1-q}=a\dfrac{1-q}{1-q}=a=aq^0=a_0$

$S_n+a_{n+1}=a\dfrac{1-q^{n+1}}{1-q}+aq^{n+1}=$

$=\dfrac{a-aq^{n+1}+aq^{n+1}-aq^{n+2}}{1-q}=\dfrac{a-aq^{n+2}}{1-q}=$

$=a\dfrac{1-q^{n+2}}{1-q}=S_{n+1}$

We can conclude that $$S_n$$ is the sequence of partial sums of $$a_n$$:

$\sum_{i=0}^{n}aq^i=a\dfrac{1-q^{n+1}}{1-q}$

### Convergence

Depending on the value of the reason we have 4 cases:

For $$q=-1$$ we have: $$\displaystyle{\sum_{n=0}^{+\infty}}a(-1)^n$$; where $$s_n=a$$ if $$n$$ is even, and $$s_n=0$$ otherwise

For $$q=1$$ we have: $$\displaystyle{\sum_{n=0}^{+\infty}}a(1)^n$$

For $$|q|<1$$ we have: $$\displaystyle{\sum_{n=0}^{+\infty}}aq^n=\dfrac{a}{1-q}$$

For $$|q|>1$$ we have: $$\displaystyle{\sum_{n=0}^{+\infty}}aq^n=+\infty$$ if $$a>0$$ and $$\displaystyle{\sum_{n=0}^{+\infty}}aq^n=-\infty$$ if $$a<0$$

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