Comprehensive Guide to Sequences and Series

What is a Sequence?

A sequence is an ordered list of numbers that follow a specific pattern. Each number in the sequence is called a term. We typically denote a sequence as:

a₁, a₂, a₃, ..., aₙ, ...

Where aₙ represents the nth term of the sequence.

What is a Series?

A series is the sum of the terms of a sequence. If we have a sequence {aₙ}, the corresponding series is:

a₁ + a₂ + a₃ + ... + aₙ + ...

We often use the sigma notation to represent a series:

∑aₙ = a₁ + a₂ + a₃ + ... + aₙ + ...

Finite and Infinite Sequences/Series

Finite sequence/series: Has a fixed number of terms.

Infinite sequence/series: Continues indefinitely.

Example:

Finite sequence: 2, 4, 6, 8, 10 (5 terms)

Infinite sequence: 2, 4, 6, 8, 10, ... (continues indefinitely)

Finite series: 2 + 4 + 6 + 8 + 10 = 30

Infinite series: 2 + 4 + 6 + 8 + 10 + ...

Arithmetic Sequence

An arithmetic sequence is a sequence where each term differs from the preceding term by a constant value called the common difference (d).

General form: a, a+d, a+2d, a+3d, ...

nth term formula: aₙ = a₁ + (n-1)d

Where:

a₁ = first term

d = common difference

n = position of the term

Example 1:

Find the 20th term of the arithmetic sequence: 5, 8, 11, 14, ...

First, identify a₁ and d:

a₁ = 5

d = 8 - 5 = 3

Using the formula aₙ = a₁ + (n-1)d:

a₂₀ = 5 + (20-1)3 = 5 + 57 = 62

Therefore, the 20th term is 62.

Example 2:

Find the arithmetic sequence with a₁ = 4 and a₁₀ = 31.

Using the formula aₙ = a₁ + (n-1)d:

a₁₀ = 4 + (10-1)d = 4 + 9d = 31

Solving for d: 9d = 27, so d = 3

The sequence is: 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, ...

Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence.

Sum of n terms: Sₙ = n/2[2a₁ + (n-1)d] = n/2[a₁ + aₙ]

Where:

a₁ = first term

aₙ = nth term

d = common difference

n = number of terms

Example 3:

Find the sum of the first 30 terms of the arithmetic sequence: 3, 7, 11, 15, ...

First, identify a₁ and d:

a₁ = 3

d = 7 - 3 = 4

Using the formula aₙ = a₁ + (n-1)d, find a₃₀:

a₃₀ = 3 + (30-1)4 = 3 + 116 = 119

Now use the sum formula: Sₙ = n/2[a₁ + aₙ]

S₃₀ = 30/2[3 + 119] = 15 × 122 = 1830

Therefore, the sum of the first 30 terms is 1830.

Key Properties:

1. The sum of evenly spaced terms of an arithmetic sequence is equal to the average of these terms multiplied by the number of terms.

2. If a sequence has terms a, b, c that are in arithmetic progression, then b = (a + c)/2.

Geometric Sequence

A geometric sequence is a sequence where each term is found by multiplying the previous term by a fixed non-zero number called the common ratio (r).

General form: a, ar, ar², ar³, ...

nth term formula: aₙ = a₁r^(n-1)

Where:

a₁ = first term

r = common ratio

n = position of the term

Example 1:

Find the 7th term of the geometric sequence: 3, 6, 12, 24, ...

First, identify a₁ and r:

a₁ = 3

r = 6/3 = 2

Using the formula aₙ = a₁r^(n-1):

a₇ = 3 × 2^(7-1) = 3 × 2^6 = 3 × 64 = 192

Therefore, the 7th term is 192.

Example 2:

Find the geometric sequence with a₁ = 5 and a₄ = 405.

Using the formula aₙ = a₁r^(n-1):

a₄ = 5r^(4-1) = 5r³ = 405

Solving for r: 5r³ = 405, so r³ = 81, thus r = 3

The sequence is: 5, 15, 45, 135, 405, ...

Geometric Series

A geometric series is the sum of the terms of a geometric sequence.

Sum of n terms: Sₙ = a₁(1-r^n)/(1-r) for r ≠ 1

Sum of infinite geometric series: S = a₁/(1-r) for |r| < 1

Where:

a₁ = first term

r = common ratio

n = number of terms

Example 3:

Find the sum of the first 6 terms of the geometric sequence: 4, 12, 36, ...

First, identify a₁ and r:

a₁ = 4

r = 12/4 = 3

Using the formula Sₙ = a₁(1-r^n)/(1-r):

S₆ = 4(1-3^6)/(1-3) = 4(1-729)/(-2) = 4 × 728/2 = 1456

Therefore, the sum of the first 6 terms is 1456.

Example 4:

Find the sum of the infinite geometric series: 16 + 4 + 1 + 1/4 + ...

First, identify a₁ and r:

a₁ = 16

r = 4/16 = 1/4

Since |r| = |1/4| = 1/4 < 1, the series converges.

Using the formula S = a₁/(1-r):

S = 16/(1-1/4) = 16/(3/4) = 16 × 4/3 = 64/3 ≈ 21.33

Therefore, the sum of the infinite series is 64/3.

Key Properties:

1. A geometric series converges only if |r| < 1.

2. If a sequence has terms a, b, c that are in geometric progression, then b² = ac.

3. The sum of an infinite geometric series with |r| ≥ 1 is undefined (the series diverges).

Harmonic Sequence

A harmonic sequence is a sequence where the reciprocals of the terms form an arithmetic sequence.

General form: If {bₙ} is an arithmetic sequence, then {1/bₙ} is a harmonic sequence.

Example 1:

The sequence 1, 1/2, 1/3, 1/4, ... is a harmonic sequence because its reciprocals (1, 2, 3, 4, ...) form an arithmetic sequence with common difference 1.

Example 2:

Show that 4, 6, 12 form a harmonic sequence.

For a harmonic sequence, the reciprocals should form an arithmetic sequence.

1/4 = 0.25, 1/6 = 0.167, 1/12 = 0.083

Check if they form an arithmetic sequence:

Second term - First term: 0.167 - 0.25 = -0.083

Third term - Second term: 0.083 - 0.167 = -0.084 ≈ -0.083

The differences are approximately equal, so 4, 6, 12 form a harmonic sequence.

Harmonic Series

The harmonic series is the sum of the reciprocals of the natural numbers:

1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ...

Important Note:

The harmonic series diverges, which means its sum approaches infinity as the number of terms increases. This is a famous result in mathematics.

Example 3:

Harmonic Mean

The harmonic mean of n numbers a₁, a₂, ..., aₙ is defined as:

H = n / (1/a₁ + 1/a₂ + ... + 1/aₙ)

Find the harmonic mean of 4, 6, and 12.

H = 3 / (1/4 + 1/6 + 1/12)

H = 3 / (3/12 + 2/12 + 1/12)

H = 3 / (6/12)

H = 3 × 12/6 = 6

Therefore, the harmonic mean is 6.

Fibonacci Sequence

The Fibonacci sequence is a famous sequence where each term is the sum of the two preceding ones, starting from 0 and 1.

General form: 0, 1, 1, 2, 3, 5, 8, 13, 21, ...

Recursive formula: Fₙ = Fₙ₋₁ + Fₙ₋₂, with F₀ = 0 and F₁ = 1

Explicit formula (Binet's formula): Fₙ = (φⁿ - (1-φ)ⁿ)/√5, where φ = (1+√5)/2 ≈ 1.618 (golden ratio)

Example:

Find the 8th Fibonacci number.

The Fibonacci sequence begins: 0, 1, 1, 2, 3, 5, 8, 13, ...

Therefore, the 8th Fibonacci number is 13.

Alternatively, using Binet's formula (though more complex for hand calculations):

F₈ = ((1+√5)/2)^8 - ((1-√5)/2)^8)/√5 ≈ 13

Properties of Fibonacci Sequence:

1. The ratio of consecutive Fibonacci numbers approaches the golden ratio φ ≈ 1.618 as n increases.

2. Every third Fibonacci number is divisible by 2.

3. Every fourth Fibonacci number is divisible by 3.

4. Every fifth Fibonacci number is divisible by 5, and so on.

Other Special Sequences

1. Triangular Numbers

The sequence of triangular numbers: 1, 3, 6, 10, 15, 21, ...

The nth triangular number: T(n) = n(n+1)/2

The 5th triangular number is T(5) = 5(5+1)/2 = 15

2. Square Numbers

The sequence of square numbers: 1, 4, 9, 16, 25, ...

The nth square number: S(n) = n²

3. Lucas Numbers

Similar to Fibonacci but starts with 2, 1 instead of 0, 1: 2, 1, 3, 4, 7, 11, 18, ...

Lₙ = Lₙ₋₁ + Lₙ₋₂, with L₁ = 2 and L₂ = 1

4. Catalan Numbers

Important in combinatorics: 1, 1, 2, 5, 14, 42, 132, ...

C(n) = (1/(n+1)) × (2n choose n) = (2n)!/((n+1)!n!)

Convergence and Divergence

A series is said to converge if the sum of its terms approaches a finite value as the number of terms increases. Otherwise, it diverges.

Tests for Convergence

1. Geometric Series Test

The series ∑ar^(n-1) from n=1 to ∞:

- Converges to a/(1-r) if |r| < 1

- Diverges if |r| ≥ 1

Example:

Determine if the series ∑(3/4)^n from n=0 to ∞ converges or diverges.

This is a geometric series with a = 1 and r = 3/4.

Since |r| = 3/4 < 1, the series converges.

The sum is a/(1-r) = 1/(1-3/4) = 1/(1/4) = 4.

2. p-Series Test

The series ∑1/n^p from n=1 to ∞:

- Converges if p > 1

- Diverges if p ≤ 1

Example:

Determine if the series ∑1/n² from n=1 to ∞ converges or diverges.

This is a p-series with p = 2.

Since p = 2 > 1, the series converges.

In fact, this specific series converges to π²/6.

3. Ratio Test

For a series ∑aₙ, calculate L = lim(n→∞) |aₙ₊₁/aₙ|:

- If L < 1, the series converges absolutely

- If L > 1, the series diverges

- If L = 1, the test is inconclusive

Example:

Use the ratio test to determine if the series ∑n!/n^n from n=1 to ∞ converges or diverges.

Let aₙ = n!/n^n

aₙ₊₁ = (n+1)!/(n+1)^(n+1) = (n+1)×n!/(n+1)^(n+1)

|aₙ₊₁/aₙ| = |(n+1)×n!/(n+1)^(n+1)| ÷ |n!/n^n|

= |(n+1)×n^n/(n+1)^(n+1)|

= |n^n/(n+1)^n|

= |(n/(n+1))^n|

Taking the limit as n approaches infinity:

L = lim(n→∞) |(n/(n+1))^n| = lim(n→∞) |(1/(1+1/n))^n| = 1/e ≈ 0.368

Since L = 1/e < 1, the series converges.

4. Comparison Tests

Direct Comparison Test: If 0 ≤ aₙ ≤ bₙ for all n ≥ N:

- If ∑bₙ converges, then ∑aₙ converges

- If ∑aₙ diverges, then ∑bₙ diverges

Limit Comparison Test: If aₙ, bₙ > 0 and lim(n→∞) aₙ/bₙ = c where c is finite and positive:

- Both series either both converge or both diverge

Example:

Use the comparison test to determine if the series ∑1/(n²+1) from n=1 to ∞ converges or diverges.

Note that for all n ≥ 1, n²+1 > n², so 1/(n²+1) < 1/n².

We know that ∑1/n² converges (p-series with p = 2 > 1).

By the direct comparison test, since ∑1/n² converges and 1/(n²+1) < 1/n², the series ∑1/(n²+1) also converges.

5. Integral Test

If f is positive, continuous, and decreasing for x ≥ 1, and aₙ = f(n), then ∑aₙ converges if and only if the improper integral ∫f(x)dx from 1 to ∞ converges.

Example:

Use the integral test to determine if the series ∑1/n from n=1 to ∞ converges or diverges.

Let f(x) = 1/x, which is positive, continuous, and decreasing for x ≥ 1.

Calculate the improper integral: ∫(1/x)dx from 1 to ∞ = lim(t→∞)[ln|x|]₁^t = lim(t→∞)[ln(t) - ln(1)] = lim(t→∞)ln(t) = ∞

Since the integral diverges, the series ∑1/n (the harmonic series) also diverges.

Common Convergent Series:

1. Geometric series with |r| < 1

2. p-series with p > 1

3. ∑1/n² (converges to π²/6)

4. ∑1/n! (converges to e - 1)

Common Divergent Series:

1. Harmonic series: ∑1/n

2. ∑1/n^p where p ≤ 1

3. ∑n

Power Series

A power series is a series of the form:

∑aₙ(x-c)^n = a₀ + a₁(x-c) + a₂(x-c)² + a₃(x-c)³ + ...

Where:

- x is the variable

- aₙ are the coefficients

- c is the center of the series

Radius and Interval of Convergence

For a power series, there exists a number R (which could be 0, a positive number, or ∞) called the radius of convergence such that:

- The series converges absolutely for |x-c| < R

- The series diverges for |x-c| > R

- For |x-c| = R, the series may converge or diverge, requiring further testing

The interval of convergence is the set of x-values for which the series converges. It could be:

(c-R, c+R), [c-R, c+R), (c-R, c+R], or [c-R, c+R]

depending on whether the series converges at the endpoints.

Example 1:

Find the radius and interval of convergence for the power series ∑(x-2)^n/n! from n=0 to ∞.

We can use the ratio test:

aₙ = (x-2)^n/n!

aₙ₊₁ = (x-2)^(n+1)/(n+1)!

|aₙ₊₁/aₙ| = |(x-2)^(n+1)/(n+1)!| ÷ |(x-2)^n/n!| = |x-2|/(n+1)

Taking the limit as n approaches infinity:

L = lim(n→∞) |x-2|/(n+1) = 0 < 1 for any value of x

Since the limit is 0 (which is less than 1) regardless of the value of x, the radius of convergence is R = ∞.

This means the interval of convergence is (-∞, ∞), i.e., the series converges for all real values of x.

Example 2:

Find the radius and interval of convergence for the power series ∑n(x-3)^n from n=1 to ∞.

We use the ratio test:

aₙ = n(x-3)^n

aₙ₊₁ = (n+1)(x-3)^(n+1)

|aₙ₊₁/aₙ| = |(n+1)(x-3)^(n+1)| ÷ |n(x-3)^n| = |(n+1)/n| × |x-3| = (1 + 1/n)|x-3|

Taking the limit as n approaches infinity:

L = lim(n→∞) (1 + 1/n)|x-3| = |x-3|

For the series to converge, we need |x-3| < 1, which gives us 2 < x < 4.

Therefore, the radius of convergence is R = 1.

Now we need to check the endpoints:

At x = 2: The series becomes ∑n(-1)^n. This is a divergent series.

At x = 4: The series becomes ∑n(1)^n = ∑n. This clearly diverges.

Therefore, the interval of convergence is (2, 4).

Operations on Power Series

If the power series for f(x) and g(x) both converge on an interval I, then:

1. Addition/Subtraction: The power series for f(x) ± g(x) converges on I.

2. Multiplication: The power series for f(x) × g(x) converges on I.

3. Division: If g(x) ≠ 0 on I, the power series for f(x)/g(x) converges on I.

4. Differentiation/Integration: If a power series has radius of convergence R, then its term-by-term derivative and integral have the same radius of convergence.

Power series are very useful for representing functions. Many important functions can be expressed as power series, including e^x, sin(x), cos(x), ln(1+x), etc.

Taylor Series

A Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It's a specific type of power series.

f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ...

or more compactly:

f(x) = ∑f^(n)(a)(x-a)^n/n! from n=0 to ∞

where f^(n)(a) is the nth derivative of f evaluated at x = a.

Maclaurin Series

A Maclaurin series is simply a Taylor series centered at a = 0:

f(x) = f(0) + f'(0)x/1! + f''(0)x²/2! + f'''(0)x³/3! + ...

or more compactly:

f(x) = ∑f^(n)(0)x^n/n! from n=0 to ∞

Common Maclaurin Series

e^x = 1 + x + x²/2! + x³/3! + ... = ∑x^n/n! (convergent for all x)

sin(x) = x - x³/3! + x⁵/5! - ... = ∑(-1)^n x^(2n+1)/(2n+1)! (convergent for all x)

cos(x) = 1 - x²/2! + x⁴/4! - ... = ∑(-1)^n x^(2n)/(2n)! (convergent for all x)

ln(1+x) = x - x²/2 + x³/3 - ... = ∑(-1)^(n+1) x^n/n (convergent for -1 < x ≤ 1)

1/(1-x) = 1 + x + x² + x³ + ... = ∑x^n (convergent for |x| < 1)

Example 1:

Find the Taylor series for f(x) = e^x centered at a = 2.

We need to find the derivatives of f(x) = e^x and evaluate them at x = 2:

f(x) = e^x, f(2) = e²

f'(x) = e^x, f'(2) = e²

f''(x) = e^x, f''(2) = e²

In fact, all derivatives of e^x are e^x, so f^(n)(2) = e² for all n.

Using the Taylor series formula:

e^x = e² + e²(x-2)/1! + e²(x-2)²/2! + e²(x-2)³/3! + ...

= e² × [1 + (x-2)/1! + (x-2)²/2! + (x-2)³/3! + ...]

= e² × e^(x-2)

This makes sense because e^x = e² × e^(x-2) by the laws of exponents.

Example 2:

Find the first four terms of the Maclaurin series for f(x) = ln(1+x) and its radius of convergence.

We need to find the derivatives of f(x) = ln(1+x) and evaluate them at x = 0:

f(x) = ln(1+x), f(0) = ln(1) = 0

f'(x) = 1/(1+x), f'(0) = 1

f''(x) = -1/(1+x)², f''(0) = -1

f'''(x) = 2/(1+x)³, f'''(0) = 2

Using the Maclaurin series formula:

ln(1+x) = 0 + 1·x/1! + (-1)·x²/2! + 2·x³/3! + ...

= x - x²/2 + x³/3 + ...

The radius of convergence can be found using the ratio test:

aₙ = (-1)^(n+1) x^n/n

|aₙ₊₁/aₙ| = |(-1)^(n+2) x^(n+1)/(n+1)| ÷ |(-1)^(n+1) x^n/n| = |x|·n/(n+1)

Taking the limit as n approaches infinity:

L = lim(n→∞) |x|·n/(n+1) = |x|·lim(n→∞) n/(n+1) = |x|·1 = |x|

For convergence, we need |x| < 1, so the radius of convergence is R = 1.

Example 3:

Use Taylor series to approximate e^0.1 with an error less than 0.001.

We use the Maclaurin series for e^x: e^x = 1 + x + x²/2! + x³/3! + ...

For x = 0.1: e^0.1 = 1 + 0.1 + (0.1)²/2! + (0.1)³/3! + ...

= 1 + 0.1 + 0.005 + 0.000166... + ...

Using the alternating series estimation theorem, the error after n terms is less than the absolute value of the (n+1)th term.

We need the error < 0.001, so we need |(0.1)^n/n!| < 0.001

For n = 3: |(0.1)⁴/4!| = |0.0001/24| ≈ 0.0000042 < 0.001

Therefore, using the first 4 terms: e^0.1 ≈ 1 + 0.1 + 0.005 + 0.000166... ≈ 1.105

The actual value of e^0.1 ≈ 1.10517, so our approximation is accurate to within 0.001.

Applications of Taylor Series:

1. Approximating complex functions with simpler polynomial expressions

2. Solving differential equations

3. Numerical integration

4. Error analysis

Common Techniques

1. Identify the Pattern

Look for common differences (arithmetic), common ratios (geometric), recursion patterns, etc.

Example:

Find the next term in the sequence: 3, 7, 11, 15, ...

Calculate the differences: 7-3=4, 11-7=4, 15-11=4

This is an arithmetic sequence with common difference d = 4.

The next term would be 15 + 4 = 19.

2. Find the General Term (Explicit Formula)

For arithmetic sequences: aₙ = a₁ + (n-1)d

For geometric sequences: aₙ = a₁r^(n-1)

For quadratic sequences: Look for second differences

Example:

Find the general term for the sequence: 2, 6, 12, 20, 30, ...

First differences: 4, 6, 8, 10, ...

Second differences: 2, 2, 2, ...

Since the second differences are constant, this is a quadratic sequence.

Try formula: aₙ = an² + bn + c

Using the first three terms to create a system of equations:

a₁ = 2: a(1)² + b(1) + c = 2 → a + b + c = 2

a₂ = 6: a(2)² + b(2) + c = 6 → 4a + 2b + c = 6

a₃ = 12: a(3)² + b(3) + c = 12 → 9a + 3b + c = 12

Solving this system: a = 1, b = 0, c = 1

Therefore, aₙ = n² + 1

Let's verify: a₄ = 4² + 1 = 17, but the actual 4th term is 20. So our guess was wrong.

Let's try aₙ = an² + bn + c again, making sure we solve the system correctly:

From the three equations, we get a = 1, b = 1, c = 0

Therefore, aₙ = n² + n

Let's verify: a₄ = 4² + 4 = 20 ✓, a₅ = 5² + 5 = 30 ✓

3. Using Recursive Formulas

Express each term as a function of previous terms.

Example:

Find a recursive formula for the sequence: 3, 5, 9, 17, 33, ...

Let's try to find a pattern:

a₂/a₁ = 5/3 ≈ 1.67

a₃/a₂ = 9/5 = 1.8

a₄/a₃ = 17/9 ≈ 1.89

a₅/a₄ = 33/17 ≈ 1.94

This doesn't seem to follow a simple pattern.

Let's try differences:

5-3=2, 9-5=4, 17-9=8, 33-17=16

The differences form a geometric sequence with common ratio 2.

Let's check if aₙ₊₁ - aₙ = 2(aₙ - aₙ₋₁):

a₃ - a₂ = 9 - 5 = 4 = 2(5 - 3) = 2(a₂ - a₁) ✓

a₄ - a₃ = 17 - 9 = 8 = 2(9 - 5) = 2(a₃ - a₂) ✓

Therefore, the recursive formula is: aₙ₊₁ = aₙ + 2(aₙ - aₙ₋₁) = 2aₙ - aₙ₋₁, with a₁ = 3 and a₂ = 5

4. Using Partial Sums

For finding the sum of the first n terms of a sequence.

Example:

Find a formula for the sum of the first n even numbers.

The sequence of even numbers is: 2, 4, 6, 8, ...

This is an arithmetic sequence with a₁ = 2 and d = 2.

The nth term is aₙ = 2 + (n-1)(2) = 2n

Using the formula for the sum of an arithmetic sequence:

Sₙ = n/2(a₁ + aₙ) = n/2(2 + 2n) = n/2(2 + 2n) = n(1 + n) = n + n²

5. Method of Differences

For finding patterns in complex sequences.

Example:

Find the next term in the sequence: 1, 4, 10, 20, 35, ...

First differences: 3, 6, 10, 15, ...

Second differences: 3, 4, 5, ...

Third differences: 1, 1, ...

Since the third differences are constant, this is a cubic sequence.

To find the next term, we continue the pattern of differences:

Next third difference = 1

Next second difference = 5 + 1 = 6

Next first difference = 15 + 6 = 21

Next term = 35 + 21 = 56

6. Using Known Formulas

Recognize sequences that match known patterns.

Example:

Find the sum: 1 + 3 + 6 + 10 + ... + 55

This is the sequence of triangular numbers: T(n) = n(n+1)/2

The last term is 55 = T(10) = 10(11)/2

So we need to find: T(1) + T(2) + T(3) + ... + T(10)

Using the formula for the sum of triangular numbers: ∑T(k) from k=1 to n = n(n+1)(n+2)/6

∑T(k) from k=1 to 10 = 10(11)(12)/6 = 220

Real-world Applications

1. Finance and Economics

Compound Interest: The formula A = P(1 + r)^n uses geometric sequences.

Present Value of Annuities: PV = PMT × (1 - (1 + r)^(-n))/r

Future Value of Annuities: FV = PMT × ((1 + r)^n - 1)/r

Example:

Calculate the amount in an account after 5 years if $1000 is deposited initially and the interest rate is 6% compounded annually.

Using the compound interest formula: A = P(1 + r)^n

A = 1000(1 + 0.06)^5 = 1000 × 1.3382 = $1,338.20

2. Physics

Kinematics: Position, velocity, and acceleration use sequences.

Wave theory: Fourier series represent periodic functions as sums of sines and cosines.

Quantum mechanics: Wave functions are expanded in series.

Example:

A ball is dropped from a height of 20 meters. Each time it hits the ground, it rebounds to 4/5 of its previous height. How far does the ball travel before coming to rest?

The ball falls 20 m, then rebounds to 20 × (4/5) = 16 m, then falls 16 m, and so on.

Distances traveled: 20 + 16 + 16 + 12.8 + 12.8 + ...

This is 20 + 2(16 + 12.8 + 10.24 + ...)

= 20 + 2 × 16(1 + 0.8 + 0.8² + ...)

= 20 + 2 × 16 × (1/(1-0.8))

= 20 + 2 × 16 × 5

= 20 + 160 = 180 meters

3. Computer Science

Algorithm analysis: Series are used to analyze time complexity.

Recursion: Many recursive algorithms follow sequences.

Computer graphics: Series are used for approximating curves.

Example:

The time complexity of the binary search algorithm is O(log n). How many comparisons are needed in the worst case to find an element in a sorted array of 1,000,000 elements?

The worst-case number of comparisons for binary search is ⌊log₂(n)⌋ + 1

For n = 1,000,000:

⌊log₂(1,000,000)⌋ + 1 = ⌊19.93⌋ + 1 = 19 + 1 = 20 comparisons