sum

Sum ( $\sum$)

$$\sum _{k=m}^n{a}_k:=a_m+a_{m+1}+\cdots +a_n$$

$k$ … index
$m,n$ … lower/upper bounds
$a_k$ … summand
Empty sum ($n<m$): $\sum :=0$, the neutral element of addition.

Where does the $\sum$ stop?

The $\sum$ binds to the product of factors immediately right of it and stops at the first top-level $+$ or $-$:

$$\sum _k{a}_k{b}_k-c=\left(\sum _k{a}_k{b}_k\right)-c$$

Multiplication and division stay inside; the first $+$/$-$ ends the scope. Parenthesise to override: ${\sum }_k(a_k+b_k)\ne {\sum }_k{a}_k+b_k$.

Infinite sums: series

Calculation rules:

Linearity

$$\sum _k(a_k+b_k)=\sum _k{a}_k+\sum _k{b}_k\sum _{k}c{a}_k=c\sum _k{a}_k$$

Sum of a constant

$$\sum _{k=m}^{n}c=(n-m+1)c$$

The range $\{m,\dots ,n\}$ has $n-m+1$ terms (inclusive). With $c=1$ this just counts them.

Splitting the range:

$$\sum _{k=m}^n{a}_k=\sum _{k=m}^p{a}_k+\sum _{k=p+1}^n{a}_k(m\le p<n)$$

Break the sum at any point. With $m=1$, a block ${\sum }_{k=p+1}^n{a}_k=s_n-s_p$ is a difference of partial sums: the identity behind the cauchy criterion for series.

Index shift (reindex):

$$\sum _{k=m}^n{a}_k=\sum _{k=m+j}^{n+j}{a}_{k-j}$$

Move the bounds by $j$, compensate inside. Aligns two sums so terms cancel (e.g. the geometric series proof).

telescoping

$$\sum _{k=m}^n(b_k-b_{k+1})=b_m-b_{n+1}$$

Neighbours cancel, only the ends survive.

Double sums

A finite double sum is a reordering of the same terms, so the order of summation can be swapped freely:

$$\sum _i\sum _j{a}_{ij}=\sum _j\sum _i{a}_{ij}$$

(For infinite sums this needs absolute convergence.)

${\sum }_{k=m}^{n}c=?$

$(n-m+1)c$. The index set $\{m,\dots ,n\}$ has $n-m+1$ elements.

Split ${\sum }_{k=m}^n{a}_k$ into the sum of two sums at a point $p$.

${\sum }_{k=m}^p{a}_k+{\sum }_{k=p+1}^n{a}_k(m\le p<n)$

Split ${\sum }_{k=n+1}^m{a}_k$ into the difference of two sums.

$$\sum _{k=n+1}^m=s_m-s_n=\sum _{k=1}^m{a}_k-\sum _{k=1}^n{a}_k$$

Shift the index of ${\sum }_{k=m}^n{a}_k$ by $j$.

${\sum }_{k=m+j}^{n+j}{a}_{k-j}$