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Lesson 3: Summation Notation and Series

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Learning Objectives

Understand summation notation (sigma notation) and its components
Write series using summation notation
Expand summation notation to write out terms
Evaluate finite sums using summation formulas
Apply formulas for arithmetic series
Apply formulas for geometric series
Use properties of summation (linearity, splitting sums)
Apply special summation formulas (sum of integers, sum of squares, sum of cubes)
Solve real-world problems using series
Evaluate infinite geometric series when they converge

Introduction: What is Summation Notation?

Summation Notation (Sigma Notation): A compact way to represent the sum of a sequence of terms using the Greek letter sigma (Σ).

General form:

Σ (from i=m to n) a_i

Components:

i is the index of summation (or counter variable)
m is the lower bound (starting value)
n is the upper bound (ending value)
a_i is the general term (expression to be summed)

The notation Σ (from i=1 to n) a_i means: a₁ + a₂ + a₃ + ... + a_n

Example 1: Understanding the Components

Identify the components of: Σ (from k=1 to 5) (2k + 3)

Solution:

Index of summation: k
Lower bound: 1
Upper bound: 5
General term: 2k + 3

Meaning: Sum the values of (2k + 3) as k goes from 1 to 5.

Example 2: Expanding Summation Notation

Write out all terms: Σ (from i=1 to 4) i²

Solution:

Step 1: Substitute each value of i from 1 to 4 into i²:

When i = 1: 1² = 1
When i = 2: 2² = 4
When i = 3: 3² = 9
When i = 4: 4² = 16

Step 2: Write as a sum:

Σ (from i=1 to 4) i² = 1 + 4 + 9 + 16 = 30

Answer: 1 + 4 + 9 + 16 = 30

Example 3: Expanding with a Starting Index Other Than 1

Expand and evaluate: Σ (from n=3 to 6) (2n - 1)

Solution:

Step 1: Substitute n = 3, 4, 5, 6:

When n = 3: 2(3) - 1 = 5
When n = 4: 2(4) - 1 = 7
When n = 5: 2(5) - 1 = 9
When n = 6: 2(6) - 1 = 11

Step 2: Sum the terms:

5 + 7 + 9 + 11 = 32

Answer: 32

Example 4: Constant Terms

Evaluate: Σ (from k=1 to 7) 5

Solution:

When the term doesn't depend on the index, we simply add that constant value multiple times.

Σ (from k=1 to 7) 5 = 5 + 5 + 5 + 5 + 5 + 5 + 5 = 7 × 5 = 35

General Rule: Σ (from i=1 to n) c = nc, where c is a constant.

Answer: 35

Example 5: Summation with Fractions

Expand and evaluate: Σ (from j=1 to 4) 1/j

Solution:

Step 1: Substitute j = 1, 2, 3, 4:

Σ (from j=1 to 4) 1/j = 1/1 + 1/2 + 1/3 + 1/4

Step 2: Find common denominator (12) and add:

= 12/12 + 6/12 + 4/12 + 3/12
= 25/12

Answer: 25/12 or approximately 2.083

Section 1: Writing Series in Summation Notation

Converting a series to summation notation requires identifying the pattern in the terms and finding a general formula.

Example 6: Writing a Simple Series

Write using summation notation: 3 + 6 + 9 + 12 + 15

Solution:

Step 1: Identify the pattern:

Terms: 3, 6, 9, 12, 15
These are multiples of 3: 3(1), 3(2), 3(3), 3(4), 3(5)

Step 2: Write the general term:

a_i = 3i

Step 3: Determine bounds:

i starts at 1 and ends at 5

Answer: Σ (from i=1 to 5) 3i

Example 7: Odd Numbers

Write using summation notation: 1 + 3 + 5 + 7 + 9 + 11

Solution:

Step 1: Identify the pattern:

These are the first 6 odd numbers
nth odd number = 2n - 1

Step 2: Verify:

n=1: 2(1)-1 = 1
n=2: 2(2)-1 = 3
n=3: 2(3)-1 = 5
n=6: 2(6)-1 = 11

Answer: Σ (from n=1 to 6) (2n - 1)

Example 8: Alternating Signs

Write using summation notation: 2 - 4 + 6 - 8 + 10

Solution:

Step 1: Notice the alternating signs and pattern in absolute values:

Absolute values: 2, 4, 6, 8, 10 (even numbers = 2k)
Signs: +, -, +, -, + (alternating)

Step 2: Use (-1)^k+1 for alternating signs starting with positive:

k=1: (-1)² = +1
k=2: (-1)³ = -1
k=3: (-1)⁴ = +1

Step 3: Combine:

General term: (-1)^k+1(2k)

Answer: Σ (from k=1 to 5) (-1)^k+1(2k)

Example 9: Squares with Alternating Signs

Write using summation notation: -1 + 4 - 9 + 16 - 25

Solution:

Step 1: Identify pattern in absolute values:

Absolute values: 1, 4, 9, 16, 25 = 1², 2², 3², 4², 5²

Step 2: Identify sign pattern:

Signs: -, +, -, +, - (alternating, starting with negative)
Use (-1)ⁿ for alternating signs starting with negative

Step 3: Verify:

n=1: (-1)¹(1²) = -1
n=2: (-1)²(2²) = +4
n=3: (-1)³(3²) = -9

Answer: Σ (from n=1 to 5) (-1)ⁿn²

Example 10: Fraction Pattern

Write using summation notation: 1/2 + 2/3 + 3/4 + 4/5 + 5/6

Solution:

Step 1: Analyze numerators and denominators:

Numerators: 1, 2, 3, 4, 5 (counting numbers)
Denominators: 2, 3, 4, 5, 6 (one more than numerator)

Step 2: Write general term:

If numerator is k, denominator is k+1
General term: k/(k+1)

Step 3: Determine bounds:

k goes from 1 to 5

Answer: Σ (from k=1 to 5) k/(k+1)

Section 2: Properties of Summation

Summation Properties

Let c be a constant and a_i, b_i be sequences. Then:

1. Constant Multiple:

Σ (from i=1 to n) c·a_i = c·Σ (from i=1 to n) a_i

2. Sum/Difference Rule:

Σ (from i=1 to n) (a_i ± b_i) = Σ (from i=1 to n) a_i ± Σ (from i=1 to n) b_i

3. Constant Sum:

Σ (from i=1 to n) c = nc

4. Splitting Sums:

Σ (from i=1 to n) a_i = Σ (from i=1 to k) a_i + Σ (from i=k+1 to n) a_i

Example 11: Using the Constant Multiple Property

If Σ (from i=1 to 10) a_i = 45, find Σ (from i=1 to 10) 3a_i

Solution:

Using the constant multiple property:

Σ (from i=1 to 10) 3a_i = 3·Σ (from i=1 to 10) a_i
= 3·(45)
= 135

Answer: 135

Example 12: Using Sum and Difference Rules

If Σ (from k=1 to 20) x_k = 100 and Σ (from k=1 to 20) y_k = 75, find Σ (from k=1 to 20) (2x_k - 3y_k)

Solution:

Step 1: Apply sum/difference and constant multiple properties:

Σ (from k=1 to 20) (2x_k - 3y_k) = Σ (from k=1 to 20) 2x_k - Σ (from k=1 to 20) 3y_k

Step 2: Factor out constants:

= 2·Σ (from k=1 to 20) x_k - 3·Σ (from k=1 to 20) y_k

Step 3: Substitute known values:

= 2(100) - 3(75)
= 200 - 225
= -25

Answer: -25

Example 13: Splitting a Sum

Evaluate: Σ (from i=1 to 5) i + Σ (from i=6 to 10) i

Solution:

This can be combined into a single sum:

Σ (from i=1 to 5) i + Σ (from i=6 to 10) i = Σ (from i=1 to 10) i

Using the formula for sum of first n integers (we'll derive this later):

Σ (from i=1 to 10) i = 10(11)/2 = 55

Or calculate directly:

(1+2+3+4+5) + (6+7+8+9+10) = 15 + 40 = 55

Answer: 55

Section 3: Special Summation Formulas

Essential Summation Formulas

1. Sum of First n Positive Integers:

Σ (from i=1 to n) i = n(n+1)/2

2. Sum of First n Squares:

Σ (from i=1 to n) i² = n(n+1)(2n+1)/6

3. Sum of First n Cubes:

Σ (from i=1 to n) i³ = [n(n+1)/2]² = [n²(n+1)²]/4

4. Constant Sum:

Σ (from i=1 to n) c = nc

Example 14: Sum of First n Integers

Evaluate: Σ (from k=1 to 50) k

Solution:

Using the formula Σ (from i=1 to n) i = n(n+1)/2 with n = 50:

Σ (from k=1 to 50) k = 50(51)/2 = 2550/2 = 1275

Answer: 1275

Example 15: Sum of First n Squares

Evaluate: Σ (from i=1 to 10) i²

Solution:

Using the formula Σ (from i=1 to n) i² = n(n+1)(2n+1)/6 with n = 10:

Σ (from i=1 to 10) i² = 10(11)(21)/6
= 2310/6
= 385

Verification: 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385

Answer: 385

Example 16: Sum of First n Cubes

Evaluate: Σ (from j=1 to 5) j³

Solution:

Using the formula Σ (from i=1 to n) i³ = [n(n+1)/2]² with n = 5:

Σ (from j=1 to 5) j³ = [5(6)/2]²
= [30/2]²
= 15²
= 225

Verification: 1³ + 2³ + 3³ + 4³ + 5³ = 1 + 8 + 27 + 64 + 125 = 225

Answer: 225

Example 17: Combining Formulas

Evaluate: Σ (from i=1 to 20) (3i² - 2i + 5)

Solution:

Step 1: Split the sum:

Σ (from i=1 to 20) (3i² - 2i + 5) = Σ (from i=1 to 20) 3i² - Σ (from i=1 to 20) 2i + Σ (from i=1 to 20) 5

Step 2: Factor out constants:

= 3·Σ (from i=1 to 20) i² - 2·Σ (from i=1 to 20) i + Σ (from i=1 to 20) 5

Step 3: Apply formulas:

Σ i² = 20(21)(41)/6 = 2870
Σ i = 20(21)/2 = 210
Σ 5 = 20(5) = 100

Step 4: Calculate:

= 3(2870) - 2(210) + 100
= 8610 - 420 + 100
= 8290

Answer: 8290

Example 18: Adjusting Bounds

Evaluate: Σ (from k=5 to 12) k

Solution:

Method 1: Use splitting property:

Σ (from k=5 to 12) k = Σ (from k=1 to 12) k - Σ (from k=1 to 4) k

= 12(13)/2 - 4(5)/2
= 78 - 10
= 68

Method 2: Direct calculation:

5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 68

Answer: 68

Section 4: Arithmetic Series

Arithmetic Series: The sum of the terms of an arithmetic sequence.

If {a_n} is an arithmetic sequence with first term a₁ and common difference d, then the sum of the first n terms is:

Sum of Arithmetic Series (Two Forms):

S_n = n(a₁ + a_n)/2

S_n = n[2a₁ + (n-1)d]/2

where a_n = a₁ + (n-1)d

Example 19: Sum of Arithmetic Series (Given First and Last Terms)

Find the sum: 3 + 7 + 11 + 15 + ... + 99

Solution:

Step 1: Identify the sequence:

a₁ = 3, d = 4, a_n = 99

Step 2: Find n using a_n = a₁ + (n-1)d:

99 = 3 + (n-1)(4)
96 = 4(n-1)
24 = n-1
n = 25

Step 3: Use S_n = n(a₁ + a_n)/2:

S₂₅ = 25(3 + 99)/2
= 25(102)/2
= 2550/2
= 1275

Answer: 1275

Example 20: Sum Using the Second Formula

Find the sum of the first 30 terms of the arithmetic sequence: 5, 9, 13, 17, ...

Solution:

Step 1: Identify values:

a₁ = 5, d = 4, n = 30

Step 2: Use S_n = n[2a₁ + (n-1)d]/2:

S₃₀ = 30[2(5) + (30-1)(4)]/2
= 30[10 + 29(4)]/2
= 30[10 + 116]/2
= 30(126)/2
= 3780/2
= 1890

Answer: 1890

Example 21: Word Problem - Seating Arrangement

An auditorium has 20 rows of seats. The first row has 25 seats, and each subsequent row has 3 more seats than the previous row. How many seats are in the auditorium?

Solution:

Step 1: This is an arithmetic series:

a₁ = 25 (seats in first row)
d = 3 (additional seats per row)
n = 20 (number of rows)

Step 2: Use the sum formula:

S₂₀ = 20[2(25) + (20-1)(3)]/2
= 20[50 + 19(3)]/2
= 20[50 + 57]/2
= 20(107)/2
= 2140/2
= 1070

Answer: The auditorium has 1070 seats.

Section 5: Geometric Series

Geometric Series: The sum of the terms of a geometric sequence.

If {a_n} is a geometric sequence with first term a₁ and common ratio r (where r ≠ 1), then the sum of the first n terms is:

Sum of Finite Geometric Series:

S_n = a₁(1 - rⁿ)/(1 - r), where r ≠ 1

Alternative form:

S_n = a₁(rⁿ - 1)/(r - 1), where r ≠ 1

Example 22: Sum of Geometric Series

Find the sum: 2 + 6 + 18 + 54 + 162

Solution:

Step 1: Identify the geometric sequence:

a₁ = 2, r = 3 (each term is 3 times the previous)
n = 5 (five terms)

Step 2: Use S_n = a₁(1 - rⁿ)/(1 - r):

S₅ = 2(1 - 3⁵)/(1 - 3)
= 2(1 - 243)/(-2)
= 2(-242)/(-2)
= -484/(-2)
= 242

Verification: 2 + 6 + 18 + 54 + 162 = 242

Answer: 242

Example 23: Geometric Series with Fractional Ratio

Find the sum of the first 6 terms: 64, 32, 16, 8, ...

Solution:

Step 1: Identify values:

a₁ = 64, r = 1/2, n = 6

Step 2: Apply the formula:

S₆ = 64[1 - (1/2)⁶]/(1 - 1/2)
= 64[1 - 1/64]/(1/2)
= 64[63/64]/(1/2)
= 63/(1/2)
= 63 × 2
= 126

Answer: 126

Example 24: Sum with Negative Ratio

Find the sum: Σ (from k=1 to 8) 3(-2)^k-1

Solution:

Step 1: Identify the geometric series:

When k=1: 3(-2)⁰ = 3(1) = 3
When k=2: 3(-2)¹ = 3(-2) = -6
So a₁ = 3, r = -2, n = 8

Step 2: Apply the formula:

S₈ = 3[1 - (-2)⁸]/(1 - (-2))
= 3[1 - 256]/(1 + 2)
= 3(-255)/3
= -255

Answer: -255

Example 25: Word Problem - Investment Growth

You invest $1000 and it grows by 5% each year. What is the total value of all annual amounts after 10 years? (Include the initial investment.)

Solution:

Step 1: This is a geometric series:

Year 1: $1000
Year 2: $1000(1.05)
Year 3: $1000(1.05)²
...
Year 10: $1000(1.05)⁹

a₁ = 1000, r = 1.05, n = 10

Step 2: Apply the formula:

S₁₀ = 1000(1 - 1.05¹⁰)/(1 - 1.05)
= 1000(1 - 1.6289)/(-0.05)
= 1000(-0.6289)/(-0.05)
≈ 12,578

Answer: The total value is approximately $12,578.

Note: This represents the sum of all yearly amounts, not the final balance.

Section 6: Infinite Geometric Series

Infinite Geometric Series: A geometric series with infinitely many terms.

The series converges (has a finite sum) only when |r| < 1.

Sum of Infinite Geometric Series:

If |r| < 1, then S = a₁/(1 - r)

If |r| ≥ 1, the series diverges (no finite sum)

Example 26: Infinite Series with r = 1/2

Find the sum: 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

Solution:

Step 1: Identify values:

a₁ = 1, r = 1/2

Step 2: Check convergence:

|r| = |1/2| = 1/2 < 1 (series converges)

Step 3: Apply formula S = a₁/(1 - r):

S = 1/(1 - 1/2)
= 1/(1/2)
= 2

Answer: 2

Example 27: Infinite Series - Repeating Decimal

Express 0.777... as a fraction using infinite geometric series.

Solution:

Step 1: Write as a series:

0.777... = 0.7 + 0.07 + 0.007 + ...
= 7/10 + 7/100 + 7/1000 + ...

Step 2: Identify the geometric series:

a₁ = 7/10, r = 1/10

Step 3: Check convergence:

|r| = 1/10 < 1

Step 4: Apply formula:

S = (7/10)/(1 - 1/10)
= (7/10)/(9/10)
= 7/9

Answer: 0.777... = 7/9

Example 28: Does the Series Converge?

Determine if the series converges: 5 + 10 + 20 + 40 + ...

Solution:

Step 1: Identify values:

a₁ = 5, r = 2

Step 2: Check convergence condition:

|r| = |2| = 2 ≥ 1

Answer: The series diverges (has no finite sum) because |r| ≥ 1. The terms keep getting larger, so the sum approaches infinity.

Check Your Understanding

1. Expand and evaluate: Σ (from k=2 to 5) (3k - 1)

Answer: 38

When k=2: 3(2)-1 = 5
When k=3: 3(3)-1 = 8
When k=4: 3(4)-1 = 11
When k=5: 3(5)-1 = 14
Sum: 5 + 8 + 11 + 14 = 38

2. Write using summation notation: 4 + 8 + 12 + 16 + 20 + 24

Answer: Σ (from i=1 to 6) 4i

Pattern: multiples of 4, so general term is 4i where i goes from 1 to 6.

3. If Σ (from i=1 to 15) a_i = 60, find Σ (from i=1 to 15) (5a_i + 2)

Answer: 330

Σ (5a_i + 2) = 5·Σ a_i + Σ 2
= 5(60) + 15(2)
= 300 + 30
= 330

4. Evaluate using a formula: Σ (from i=1 to 25) i

Answer: 325

Using Σ i = n(n+1)/2:
= 25(26)/2
= 650/2
= 325

5. Evaluate: Σ (from k=1 to 8) k²

Answer: 204

Using Σ i² = n(n+1)(2n+1)/6:
= 8(9)(17)/6
= 1224/6
= 204

6. Find the sum of the arithmetic series: 5 + 11 + 17 + 23 + ... + 101

Answer: 901

a₁ = 5, d = 6, a_n = 101
Find n: 101 = 5 + (n-1)(6), so n = 17
S₁₇ = 17(5 + 101)/2 = 17(106)/2 = 901

7. Find the sum of the geometric series: 3 + 12 + 48 + 192 + 768

Answer: 1023

a₁ = 3, r = 4, n = 5
S₅ = 3(1 - 4⁵)/(1 - 4)
= 3(1 - 1024)/(-3)
= 3(-1023)/(-3)
= 1023

8. Find the sum of the infinite geometric series: 6 + 2 + 2/3 + 2/9 + ...

Answer: 9

a₁ = 6, r = 1/3
Since |r| = 1/3 < 1, series converges
S = 6/(1 - 1/3) = 6/(2/3) = 9

9. Write using summation notation with alternating signs: 1 - 1/2 + 1/3 - 1/4 + 1/5

Answer: Σ (from n=1 to 5) (-1)ⁿ⁺¹/n

Alternating signs starting with positive: (-1)ⁿ⁺¹
Denominators: n
Combined: (-1)ⁿ⁺¹/n

10. Does the infinite series 8 + 4 + 2 + 1 + 1/2 + ... converge? If so, find its sum.

Answer: Yes, it converges to 16

a₁ = 8, r = 1/2
|r| = 1/2 < 1, so it converges
S = 8/(1 - 1/2) = 8/(1/2) = 16

Key Takeaways

Summation notation Σ provides a compact way to represent series
The index, lower bound, upper bound, and general term are essential components
Summation has linear properties: constants can be factored out and sums can be split
Special formulas exist for common series: Σ i = n(n+1)/2, Σ i² = n(n+1)(2n+1)/6, Σ i³ = [n(n+1)/2]²
Arithmetic series sum: S_n = n(a₁ + a_n)/2 or S_n = n[2a₁ + (n-1)d]/2
Geometric series sum: S_n = a₁(1 - rⁿ)/(1 - r) for finite series
Infinite geometric series converge only when |r| < 1, with sum S = a₁/(1 - r)
Alternating signs can be represented using (-1)ⁿ or (-1)ⁿ⁺¹
Complex sums can be evaluated by splitting, applying properties, and using formulas
Series have practical applications in finance, physics, and real-world problem-solving

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