Summation

Summation is the operation of adding a sequence of numbers to get a sum or total. It is usually denoted with the letter sigma ($\sum$). A sum of all the numbers from 1 to 5 can be written as:

$\sum _{i\mathop {=} 1}^{5}i=1+2+3+4+5=15$

Any operation can be performed on i. For instance,

$\sum _{i\mathop {=} 1}^{5}i^{3}=1^{3}+2^{3}+3^{3}+4^{3}+5^{3}=1+8+27+64+125=225$

A partial sum, where the sum is only of part of a series, is also called a finite sum. If a sum is between a number and infinity, it is called a series. Infinite sums sometimes approach infinity (such as $\lim _{n\to \infty }\sum _{i\mathop {=} 1}^{n}i=\infty$) and sometimes equal a specific value (for instance, $\lim _{n\to \infty }\sum _{i\mathop {=} 1}^{n}{\frac {1}{i}}=2$).

Infinite sums are divided into convergent and divergent series. With convergent series, the difference between terms decreases (as seen in $\sum _{i\mathop {=} 1}^{\infty }{\frac {1}{i}}$), whereas with divergent series, the difference increases (such as $\sum _{i\mathop {=} 1}^{\infty }i$). The limit of a convergent series is a defined point, whereas the limit of divergent series is infinite.

It is important to note that the limit, or the value the sum approaches, is not the same as what the sum equals. For instance, $\lim _{n\to \infty }\sum _{i\mathop {=} 1}^{n}i=\infty$, but $\sum _{i\mathop {=} 1}^{\infty }i=-{\frac {1}{12}}$

Properties

A sum of any series in which i is multiplied or divided by a constant is the same as the entire sum multiplied or divided by said constant. If C is a constant,

$\sum _{i\mathop {=} m}^{n}Ci=C\sum _{i\mathop {=} m}^{n}i$

If the sequence is of two numbers added to each other, the answer will be the same as the sums of both terms added together.

$\sum _{i\mathop {=} m}^{n}(i+i)=\sum _{i\mathop {=} m}^{n}i+\sum _{i\mathop {=} m}^{n}i$

Sums where $i$ is raised to a power can be found with the following formulas:

$\sum _{i\mathop {=} 1}^{n}(i^{1})={\frac {n(n+1)}{2}}$

$\sum _{i\mathop {=} 1}^{n}(i^{2})={\frac {n(n+1)(2n+1)}{6}}$

$\sum _{i\mathop {=} 1}^{n}(i^{3})={\frac {n^{2}(n+1)^{2}}{4}}$