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Infinite Series

Historical Background of Infinite Series

Problem: Calculating the Area of a Circle

Imagine you are a mathematician in the 17th century, facing a seemingly simple question: How can we accurately calculate the area of a circle?

The formula for the area of a circle is known: A=πr2A = \pi r^2

But what is the value of π? At that time, people only knew that π was about 3.14, but needed a more accurate value.

The Problem with the Geometric Method

The ancient Greek mathematician Archimedes computed π with a geometric method:

  1. Inscribe a regular hexagon in the circle and compute its area
  2. Circumscribe a regular hexagon about the circle and compute its area
  3. The area of the circle lies between these two values
  4. Increase the number of sides (regular dodecagon, regular 24-gon, …) to squeeze the value closer

Diagram: Archimedes squeezed the area of the circle between regular polygons inscribed in and circumscribed about it; adding more sides made the bound on π progressively tighter.

The problem: this method is extremely tedious to carry out, and its precision is limited.

The Geometric Series Summation Formula

One day, mathematicians noticed a remarkable phenomenon:

11−x=1+x+x2+x3+x4+⋯\frac{1}{1 - x} = 1 + x + x^2 + x^3 + x^4 + \cdots

When x=12x = \frac{1}{2}:

11−12=1+12+14+18+116+⋯\frac{1}{1 - \frac{1}{2}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots

Left-hand side: 11−12=112=2\frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2

Right-hand side: 1+12+14+18+116+⋯1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots

The discovery: this infinite sum turns out to be exactly 2!

If you are interested in the proof, here is Euler’s method:

infinite serieshistoryEuler

Approach:

Euler proved the geometric series formula by an algebraic method. The proof rests on the following observation: multiplying a series by xx shifts every term one place to the right, so subtracting the shifted series from the original leaves only the first term.

Steps:

  1. Introduce the sum function: let S=1+x+x2+x3+x4+⋯S = 1 + x + x^2 + x^3 + x^4 + \cdots
  2. Multiply by xx: xS=x+x2+x3+x4+x5+⋯xS = x + x^2 + x^3 + x^4 + x^5 + \cdots
  3. Subtract: S−xS=1S - xS = 1, because every term except the leading 11 cancels
  4. Factor out: (1−x)S=1(1 - x)S = 1
  5. Solve: S=11−xS = \frac{1}{1 - x}

Answer:

The proof is complete: 11−x=1+x+x2+x3+x4+⋯\frac{1}{1 - x} = 1 + x + x^2 + x^3 + x^4 + \cdots. The convergence condition is that the series converges when ∣x∣<1|x| < 1. Historically, this proof is concise and ingenious — it demonstrates the basic technique of manipulating infinite sums and laid the groundwork for the later development of series theory.

The Birth of Infinite Series

Mathematicians realised that an infinite sum can represent a definite number!

This is the origin of infinite series.

Applying the Formula to π

Later, the mathematician Leibniz found:

π4=1−13+15−17+19−⋯\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots

This means π can be computed by simple addition and subtraction!

infinite serieshistoryLeibniz

Approach:

Between 1673 and 1674 Leibniz discovered this formula by studying the power series expansion of the arctangent function.

Steps:

  1. Study the arctangent function: Leibniz noticed the power series expansion of arctan⁡x\arctan x: arctan⁡x=x−x33+x55−x77+⋯\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots

  2. Substitute a special value: when x=1x = 1, arctan⁡1=π4\arctan 1 = \frac{\pi}{4}, so substituting gives: π4=1−13+15−17+⋯\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots

  3. Historical significance: this was the first formula in history to express π as an infinite series, providing a new way to compute π.

Answer:

Properties of the formula: its rate of convergence is slow, so many terms are needed to get an accurate value; but the computation itself is simple, involving only addition and subtraction. It gave an important line of thought for the later computation of π.

Exercises

Exercise 1

Compute the sum of the first 4 terms of the series 1+12+14+181 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8}.

infinite serieshistory

Approach:

Compute the sum of the first 4 terms directly.

Steps:

  1. Evaluate each term: 1=11 = 1, 12=0.5\frac{1}{2} = 0.5, 14=0.25\frac{1}{4} = 0.25, 18=0.125\frac{1}{8} = 0.125
  2. Add them up: 1+0.5+0.25+0.125=1.8751 + 0.5 + 0.25 + 0.125 = 1.875

Answer:

The sum of the first 4 terms is 1.875.

Exercise 2

Use the Leibniz series to compute an approximation of π from its first 3 terms.

infinite serieshistory

Approach:

Use the formula: π4=1−13+15−⋯\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \cdots

Steps:

  1. Take the first 3 terms: 1−13+151 - \frac{1}{3} + \frac{1}{5}
  2. Evaluate: 1−0.3333+0.2=0.86671 - 0.3333 + 0.2 = 0.8667
  3. Multiply by 4: π≈4×0.8667=3.4668\pi \approx 4 \times 0.8667 = 3.4668

Answer:

π ≈ 3.4668 (approximation from the first 3 terms)

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
π\piGreek letterPiCircular constant, approximately 3.14159
∑\sumGreek letterSigmaSummation symbol, representing series
∞\inftyMathematical symbolInfinityRepresents infinite series, infinite number of terms
xxMathematical symbolVariableVariable in the series
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The Basic Concepts of Infinite Series

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