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Sigma Notation for Sums

MATH161
Reference: Stewart §4.1  •  Chapter: 4  •  Section: 1

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Sigma Notation for Sums

Writing Long Sums Without the "..."

When we approximate areas under curves, we need to add up many terms—sometimes hundreds or thousands. Writing

$$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + \cdots + f(x_n)\Delta x$$

is tedious and imprecise (what exactly do the dots mean?). Sigma notation gives us a compact, precise way to write such sums.

The symbol $\Sigma$ (Greek capital "sigma") means "sum." It's like a compact for-loop in programming: it tells you what to add, where to start, and where to stop.

Prerequisite Map

Prerequisites
Function EvaluationAlgebraic Manipulation
This skill
Sigma Notation

Quick Reference

Property Value
Section Stewart §4.1
Course MATH161
Difficulty Beginner
Time ~15 minutes

Key Concepts

The Sigma Notation

$$\sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \cdots + a_n$$

Reading it: "The sum from $i = 1$ to $n$ of $a_i$"

Components:

    n        ← upper limit (stop here)
   ___
   \
    >   a_i  ← general term
   /___
   i=1       ← index and lower limit (start here)

Expanding Sigma Notation

To expand, substitute each integer from the lower limit to the upper limit:

$$\sum_{i=1}^{4} i^2 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30$$

$$\sum_{j=0}^{3} 2^j = 2^0 + 2^1 + 2^2 + 2^3 = 1 + 2 + 4 + 8 = 15$$

Writing in Sigma Notation

To convert an explicit sum to sigma notation:

  1. Identify the pattern (what changes from term to term?)
  2. Write the general term using an index
  3. Determine the starting and ending values

Example: Write $2 + 4 + 6 + 8 + 10$ in sigma notation.

Key Summation Formulas

These formulas let us evaluate sums without expanding:

$$\boxed{\sum_{i=1}^{n} c = nc} \quad \text{(sum of constant)}$$

$$\boxed{\sum_{i=1}^{n} i = \frac{n(n+1)}{2}} \quad \text{(sum of first } n \text{ integers)}$$

$$\boxed{\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}} \quad \text{(sum of squares)}$$

$$\boxed{\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2} \quad \text{(sum of cubes)}$$

Linearity Properties

$$\sum_{i=1}^{n} (a_i + b_i) = \sum_{i=1}^{n} a_i + \sum_{i=1}^{n} b_i$$

$$\sum_{i=1}^{n} c \cdot a_i = c \cdot \sum_{i=1}^{n} a_i$$

These let you break apart complicated sums!

Connection to Riemann Sums

The sum of rectangle areas can be written compactly:

$$R_n = f(x_1)\Delta x + f(x_2)\Delta x + \cdots + f(x_n)\Delta x = \sum_{i=1}^{n} f(x_i)\Delta x$$

This is why sigma notation is essential for integration.

Practice Problems

Level 1 Expanding Sigma Notation

Expand and evaluate:

(a) $\sum_{k=1}^{5} k$

(b) $\sum_{i=2}^{4} i^2$

(c) $\sum_{j=0}^{3} (2j + 1)$

Thought Process

For each sum:

  1. Substitute the index values from lower limit to upper limit
  2. Write out each term
  3. Add them up
Show Answer

(a) $\sum_{k=1}^{5} k = 1 + 2 + 3 + 4 + 5 = 15$

(b) $\sum_{i=2}^{4} i^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29$

(c) $\sum_{j=0}^{3} (2j + 1) = (2(0)+1) + (2(1)+1) + (2(2)+1) + (2(3)+1)$ $= 1 + 3 + 5 + 7 = 16$

Level 2 Writing in Sigma Notation

Write each sum in sigma notation:

(a) $1 + 4 + 9 + 16 + 25$

(b) $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$

(c) $3 + 6 + 9 + 12 + \cdots + 30$

Thought Process

For each:

  1. Identify the pattern—what's the formula for the $i$th term?
  2. Determine starting and ending index values
  3. Write using $\sum$
Show Answer

(a) These are perfect squares: $1^2, 2^2, 3^2, 4^2, 5^2$

$$\sum_{i=1}^{5} i^2$$

(b) These are powers of $\frac{1}{2}$: $\left(\frac{1}{2}\right)^1, \left(\frac{1}{2}\right)^2, \left(\frac{1}{2}\right)^3, \left(\frac{1}{2}\right)^4$

$$\sum_{i=1}^{4} \left(\frac{1}{2}\right)^i \quad \text{or} \quad \sum_{i=1}^{4} \frac{1}{2^i}$$

(c) These are multiples of 3: $3(1), 3(2), 3(3), \ldots, 3(10)$

$$\sum_{i=1}^{10} 3i$$

Level 3 Using Summation Formulas

Evaluate without expanding:

(a) $\sum_{i=1}^{100} i$

(b) $\sum_{k=1}^{20} k^2$

(c) $\sum_{j=1}^{50} (2j - 1)$

Thought Process

Use the summation formulas! For (c), use linearity to split the sum: $$\sum (2j - 1) = 2\sum j - \sum 1$$

Show Answer

(a) Using $\sum_{i=1}^n i = \frac{n(n+1)}{2}$:

$$\sum_{i=1}^{100} i = \frac{100(101)}{2} = \frac{10100}{2} = 5050$$

(b) Using $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$:

$$\sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = \frac{17220}{6} = 2870$$

(c) Split using linearity:

$$\sum_{j=1}^{50} (2j - 1) = 2\sum_{j=1}^{50} j - \sum_{j=1}^{50} 1$$

$$= 2 \cdot \frac{50(51)}{2} - 50 = 2550 - 50 = 2500$$

Note: This is the sum of the first 50 odd numbers, which equals $50^2 = 2500$.

Level 4 Riemann Sum Setup

For $f(x) = x^2$ on $[0, 2]$ with $n$ subintervals of equal width:

(a) What is $\Delta x$?

(b) What is $x_i$ (the right endpoint of the $i$th subinterval)?

(c) Write $R_n$ (the right Riemann sum) in sigma notation.

(d) Use a summation formula to express $R_n$ as a single fraction in terms of $n$.

Thought Process
  1. $\Delta x = \frac{b-a}{n} = \frac{2-0}{n}$
  2. Right endpoints: $x_i = a + i\Delta x = \frac{2i}{n}$
  3. $R_n = \sum f(x_i)\Delta x$
  4. Substitute, simplify, use $\sum i^2$ formula
Show Answer

(a) $\Delta x = \frac{2 - 0}{n} = \frac{2}{n}$

(b) $x_i = 0 + i \cdot \frac{2}{n} = \frac{2i}{n}$

(c) $$R_n = \sum_{i=1}^{n} f(x_i) \Delta x = \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n}$$

$$= \sum_{i=1}^{n} \frac{4i^2}{n^2} \cdot \frac{2}{n} = \sum_{i=1}^{n} \frac{8i^2}{n^3}$$

(d) Factor out constants: $$R_n = \frac{8}{n^3} \sum_{i=1}^{n} i^2 = \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$$

$$= \frac{8n(n+1)(2n+1)}{6n^3} = \frac{4(n+1)(2n+1)}{3n^2}$$

Level 5 Evaluating a Limit of Riemann Sums

Using the result from Level 4, find:

$$\lim_{n \to \infty} R_n = \lim_{n \to \infty} \frac{4(n+1)(2n+1)}{3n^2}$$

What does this represent geometrically?

Thought Process

This is a limit at infinity of a rational function. Expand the numerator and compare degrees with the denominator, or divide everything by $n^2$.

Geometrically, this limit gives the exact area under $f(x) = x^2$ from $x = 0$ to $x = 2$.

Show Answer

$$\lim_{n \to \infty} \frac{4(n+1)(2n+1)}{3n^2}$$

Expand the numerator: $(n+1)(2n+1) = 2n^2 + 3n + 1$

So: $$\lim_{n \to \infty} \frac{4(2n^2 + 3n + 1)}{3n^2} = \lim_{n \to \infty} \frac{8n^2 + 12n + 4}{3n^2}$$

Divide numerator and denominator by $n^2$: $$= \lim_{n \to \infty} \frac{8 + \frac{12}{n} + \frac{4}{n^2}}{3} = \frac{8 + 0 + 0}{3} = \frac{8}{3}$$

Geometric interpretation:

This is the exact area under the curve $y = x^2$ from $x = 0$ to $x = 2$.

We can verify: $\int_0^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$ ✓

CCI-Style Conceptual Questions

Level 2 Conceptual Understanding

True or False (justify your answer):

(a) $\sum_{i=1}^{n} i$ and $\sum_{k=1}^{n} k$ represent the same sum.

(b) $\sum_{i=1}^{5} 3 = 3$

(c) $\sum_{i=1}^{n} (a_i \cdot b_i) = \left(\sum_{i=1}^{n} a_i\right) \cdot \left(\sum_{i=1}^{n} b_i\right)$

Thought Process

For (a): The index variable is just a "counter"—does changing its name change the sum?

For (b): Count how many terms are in the sum. The constant 3 is added how many times?

For (c): Try a small example with $n=2$ to test whether sums distribute over products.

Show Answer

(a) True. The index variable is a "dummy variable"—its name doesn't matter. Both sums equal $1 + 2 + 3 + \cdots + n$.

(b) False. $\sum_{i=1}^{5} 3 = 3 + 3 + 3 + 3 + 3 = 15$, not 3. When summing a constant $c$ over $n$ terms, you get $nc$.

(c) False. Sums don't distribute over products.

Counterexample: Let $a_1 = a_2 = 1$ and $b_1 = b_2 = 1$.

  • Left side: $\sum_{i=1}^{2} (a_i \cdot b_i) = 1 \cdot 1 + 1 \cdot 1 = 2$
  • Right side: $(1+1)(1+1) = 4$

These are not equal!

Mastery Checklist

Mental Model

The For-Loop Analogy:

Think of sigma notation like a programming for-loop:

sum = 0
for i = 1 to n:
    sum = sum + f(i)

This is exactly $\sum_{i=1}^{n} f(i)$. The index $i$ is just a counter that keeps track of which term we're on.


Connections

Looking back:

Looking ahead:


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Closed Interval Method Section Index Riemann Sums

Last updated: 2026-01-22