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Lecture 03

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Jakob Bernoulli & Convergence Series

An infinite series is an ordered sequence of numbers usually given as a formula representing the nth element.

Infinite series were the subject of mathematical study by the 1660s.

Convergence/Divergence

When we take the sum (or product) of a series, the series is found to converge (arrive at a limiting value) or diverge (grow without bounds).

The mathematician Cauchy stressed the importance of considering the convergent or divergent behaviour of these sums and products of series.

Numerical Series

A numerical series in \Re:

\sum_{i=0}^\infty a i = a_0 + a_1 + a_2 + a_3 + \cdots

Functional Series

And this is a functional series:

\sum_{i=0}^\infty f_i (x) = f_0 (x) + f_1 (x) + f_2 (x) + f_3 (x) + \cdots

Power Series

A power series has the form:

\sum_{n=0}^\infty a_n x^n

Geometrical Reasoning About Infinite Series

Consider the series:

\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots = 1

In this case, we have the surprising situation that inspection of a geometrical figure leads to a result about infinite series. The following figure, depicting recursive subdivision of the unit square, makes clear why this series converges to 1:

Unit-square-series

Newton's Binomial Theorem

Isaac Newton arrived at this theorem in the 1660s:

1 + a^n = 1 + n_a + \frac{n(n-1)}{2} \cdot a^2 + \frac{n(n-1)(n-2)}{3!}\cdot a^3 + \cdots

Let n = \frac{1}{2}; then this infinite series, the "Taylor polynomial", gives one the square root of an arbitrary number.

 TO DO: insert figure (unit-circle-axes)

x^2 + y^2 = 1 \rightarrow y = \sqrt{1-x^2}

A = \pi (unit circle)

\frac{\pi}{4} = \int_0^1 \sqrt{1-x^2}\,dx

Newton used the Binomial Theorem, discussed previously, to evaluate (1-x^2)^\frac{1}{2} to obtain an infinite series and then perform term-by-term integration.

Newton thus obtained an early estimate of the value of π as the area of the unit circle.

Harmonic Series

The harmonic series, so-named because the values used in the denominators are related to the note values commonly found in musical harmonies, looks like:

S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots

This infinite series was studied by Jakob Bernoulli, whose approach was to take partial sums of the series in order to see whether it approached some limited value.

  • adding 83 terms yields ≈ 5
  • adding 1200 terms yields ≈ 10
  • adding 25,000,000 terms yields ≈ 20

Does this series converge or diverge?

\frac{1}{n} \rightarrow 0 as n \rightarrow \infty, so intuition suggests that the series converges.

However, that is not the case. This series actually diverges, but very slowly…

A series converges as the nth term approaches 0, but the converse is not true. [explain further]

Leibniz, who was living in Paris and working as an ambassador in the 1670s, explained the divergence using the "triangular numbers" as follows:

 TO DO: insert figure (triangular numbers)

T = 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \cdots

Note: 1 + 2 + \cdots + k = \frac{k(k+1)}{2}

This is a "telescoping series" that evaluates to 1 because we can cancel terms, e.g. -1/2 and +1/2:


\begin{array}{lcl}
\frac{1}{2}T & = & \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \cdots \\
& = & ( 1 - \frac{1}{2} ) + ( \frac{1}{2} - \frac{1}{3} ) + ( \frac{1}{3} - \frac{1}{4} ) + \cdots \\
& = & 1 \\
\end{array}

\Rightarrow T = 2

Jakob Bernoulli published his results in Basel, Switzerland, in 1684, in a book entitled Acta Eruditorum:

T = 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \cdots = 2

C = \frac{T}{2} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \cdots = 1

By subtracting \frac{1}{2} from both sides of the equation:

D = \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \cdots = ( 1 - \frac{1}{2} ) = \frac{1}{2}

By subtracting \frac{1}{6} from both sides of the equation:

E = \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \cdots = ( \frac{1}{2} - \frac{1}{6} ) = \frac{1}{3}

By subtracting \frac{1}{12} from both sides of the equation:

F = \frac{1}{20} + \frac{1}{30} + \cdots = ( \frac{1}{3} - \frac{1}{12} ) = \frac{1}{4}

etc.

Now add the leftmost columns:


\begin{array}{lcl}
A & = & C + D + E + F + \cdots \\
& = & \frac{1}{2} + ( \frac{1}{6} + \frac{1}{6} ) + ( \frac{1}{12} + \frac{1}{12} + \frac{1}{12} ) + ( \frac{1}{20} + \frac{1}{20} + \frac{1}{20} + \frac{1}{20} ) + \cdots \\
& = & \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots \\
\end{array}

Now add the rightmost columns:


\begin{array}{lcl}
A & = & C + D + E + F + \cdots \\
& = & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots \\
& = & 1 + A
\end{array}

This implies that A = \infty (adding 1 to \infty is still \infty). We conclude that the harmonic series is divergent.

Today we would prove divergence by examining the partial sums, seeing where they trend, i.e. towards or away from zero.

The Basel Problem

Here is an example of a convergent sequence:

S = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots

Jakob Bernoulli asked the question: what is S? This was a major unsolved problem in early 18th century mathematics.

Leonhard Euler received private math lessons from the younger Bernoulli, Johann, and who achieved as his first major result the solution to this problem:

S = \frac{\pi}{4}

The subject of the first course reading is the method in which Euler solved this problem.

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