Sigma Notation — Quick Reference
Σ Notation basics
- Mathematicians use the sigma notation to denote a sum. The uppercase Greek letter Σ indicates a sum.
- Notation consists of: summand, index, lower bound, and upper bound.
- General form: \sum_{i=m}^{n} f(i) reads as “the summation of f(i) from i = m to n,” where m\le n are integers and f(i) is a term (summand).
Components of sigma notation
- Index: i
- Lower bound: m
- Upper bound: n
- Summand: f(i)
- Expansion form: \sum_{i=m}^{n} f(i) = f(m) + f(m+1) + \cdots + f(n)
Examples of expanding sums
- Example 1: \sum_{i=2}^{4} (2i - 3) = (2\cdot2 - 3) + (2\cdot3 - 3) + (2\cdot4 - 3) = 1 + 3 + 5 = 9
- Example 2: \sum_{i=0}^{5} 2^{i} = 1 + 2 + 4 + 8 + 16 + 32 = 63
Write each expression in sigma notation
- 1 + 1/2 + 1/3 + 1/4 + \cdots + 1/100 = \sum_{i=1}^{100} \frac{1}{i}
- -1 + 2 - 3 + 4 - 5 + 6 - 7 + \cdots - 25 = \sum_{i=1}^{25} (-1)^i i
- a2 + a4 + a6 + a8 + \cdots + a20 = \sum{i=1}^{10} a{2i}
- 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 = \sum_{j=0}^{7} \left(\frac{1}{2}\right)^j
Seatwork / properties of sigma notation
- Basic property: Sum of a linear combination
\sum{i=m}^{n} [c\,f(i) + d\,g(i)] = c\,\sum{i=m}^{n} f(i) + d\,\sum_{i=m}^{n} g(i) - Constant multiple rule: brought out of the summation.
- Sum of a sequence: \sum{i=1}^{n} ai = a1 + a2 + \cdots + a_n
- Common results:
- \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
- \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}
Key evaluated sums
- Example: \sum_{i=1}^{30} (4i - 5) = 1710
- Example: \sum_{k=1}^{99} \frac{1}{k(k+1)} = \frac{99}{100}
More practice: quick reference conversions
- Write the expression in sigma notation: 1 + 1/2 + 1/3 + \cdots + 1/100 → \sum_{i=1}^{100} \frac{1}{i}
- Write the alternating sum: -1 + 2 - 3 + 4 - 5 + \cdots - 25 → \sum_{i=1}^{25} (-1)^i i
- Even terms sum: a2 + a4 + a6 + a8 + a10 → \sum{i=1}^{5} a{2i}
- Sum of a finite sequence: \sum{i=1}^{n} ai = a1 + a2 + \cdots + a_n
- Linearity and constants allow combining and scaling sums.
- Common closed forms:
- \sum_{i=1}^{n} i = \frac{n(n+1)}{2}
- \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}