Arithmetic Series

An arithmetic series is of the form `\displaystyle\sum_{n=1}^ka_n=a_1+a_2+a_3+\ldots`
where `a_{n+1}=a_n+d` for some constant `d`, called the common difference.

Gauss famously demonstrated the sum of integers with the following argument:
`\text{Let }S=\displaystyle\sum_{k=1}^nk`

`\begin{aligned} S=1+2+3+\ldots+(n-2)+(n-1)+n \\ S=n+(n-1)+(n-2)+\ldots+3+2+1 \end{aligned}`
`2S=(n+1)+(n+1)+(n+1)+\ldots+(n+1)+(n+1)+(n+1) \pod {n\text{ terms}}`

`\displaystyle 2S=n(n+1) \implies S=\frac{n(n+1)}2`

Using induction, it can be proven that, for an arithmetic sequence / series, `a_n=a_1+d(n-1)`
As a result, the sum `S` of a finite arithmetic series can be written as follows:

`\displaystyle S=\sum_{i=1}^na_i=\sum_{i=1}^n{(a_1+d(i-1))}=a_0*n+d\sum_{i=1}^n(i-1)=a_1*n+d*(\frac{n(n+1)}2-n)`
`\displaystyle =n(a_1+\frac{d(n+1)}2-d)=n(\frac{a_1}2+\frac{a_1+d(n-1)}2)=n(\frac{a_1+a_n}2)`

`\therefore \boxed{\sum_{i=1}^na_i=n(\frac{a_1+a_n}2)}`

By the divergence test, it is immediately obvious that an infinite arithmetic series diverges.

Geometric Series

A geometric series is of the form `\displaystyle\sum_{n=1}^ka_n=a_1+a_2+a_3+\ldots`
where `a_{n+1}=a_n\cdot r` for some constant `r`, called the common ratio.

Important Note: It is clear that such a sequence / series will converge if and only if `\lvert r \rvert< 1`

Using induction, it can be proven that, for a geometric sequence / series, `a_n=a_0\cdot r^n`
For an infinite geometric series,
`\displaystyle S=\sum_{i=0}^\infty a_i`

`S=a_0+(a_0\cdot r)+(a_0\cdot r^2)+(a_0\cdot r^3)+\ldots`
`S\cdot r=(a_0\cdot r)+(a_0\cdot r^2)+(a_0\cdot r^3)+\ldots`

`\displaystyle S(1-r)=a_0 \implies S=\boxed{\sum_{i=0}^\infty a_i=\frac{a_0}{1-r}}`

For a finite geometric series,
`\displaystyle S=\sum_{i=0}^n a_i=(\sum_{i=0}^\infty a_i-\sum_{i=n+1}^\infty a_i)=(\frac{a_0}{1-r}-\frac{a_0\cdot r^{n+1}}{1-r})`

`\therefore \boxed{\sum_{i=0}^na_i=a_0(\frac{1-r^{n+1}}{1-r})} \iff \boxed{\sum_{i=1}^na_i=a_1(\frac{1-r^n}{1-r})}`