Two very interesting Series
04 Nov 2019Table of contents
I just want to give two very interesting cases of series a whirl. But first, I need to introduce some context:
Definitions
We call a set of numbers \(\lbrace a_1, a_2, \dots, a_n \rbrace\) a sequence. We can say a sequence is indexed by a subset \(J\) of the natural numbers \(\mathbb{N}\) if we can describe the set above as1
\[\lbrace a_i \vert \text{where }i\in J \rbrace\]And we will call \(J\) the indexing set. In the case where the number of elements in \(J\) (sometimes written \(\vert J \vert\), read cardinality of \(J\)) is infinite, by convention, we will write \(J = \mathbb{N}\), we can call that sequence an infinite sequence.
If we have a sequence \(\lbrace a_1, a_2, \dots, a_n \rbrace\) with an indexing set \(J\), we sometimes write the sequence as:
\[\lbrace a_i \rbrace_{i\in J}\]and this way we can write both the sequence elements and the indexing set in one go.
We call a set countably infinite if there is a possible one-to-one mapping from it to the natural numbers \(\mathbb{N}\). We call it uncountably infinite, or for a much nicer name, we say that it has the cardinality of the continuum if there is no such mapping.
We can sum all the elements in a sequence, we will write this sum as:
\[\sum_{i\in J} a_i\]in a general sense, but there are two other cases, if \(J\) is a sequence of natural numbers from 1 to \(n\), we can write:
\[\sum_{i=1}^n a_i\]or, if \(J= \mathbb{N}\):
\[\sum_{i=1}^\infty a_i\]and yes, 0 is not a natural number. When \(J\) is countably infinite, we call this a series
Types of series
Basically, we can either have a series that will sum up to \(\infty\) or one that sums up to a finite amount (we can also have one type that just oscilates back and forth, but that’s not important now).
We will call the first type divergent and the second type convergent.
Now the two interesting examples:
The Geometric Series
Most people have seen this one, but I like the proof, specially, using it to remember the formula.
We define recursively a geometric series as (indexing starting at 0, so \(J = \mathbb{N}\cup\lbrace 0 \rbrace\), where \(\cup\) means “joined”):
\[\begin{eqnarray} a_0 &=& 1 \\ a_{i+1} &=& a_i\cdot r \end{eqnarray} \qquad\text{where }r\text{ is a real constant called ratio}\]If \(r\ge 1\), we can check quickly that the sum cannot converge, as the values keep getting bigger, or, better put:\(\begin{eqnarray} a_0 &=& 1 \\ a_{i+1} &=& a_i\cdot r \end{eqnarray} \qquad\text{where }r\text{ is a real constant}\)
\[\lim_{i\to\infty} a_i =\infty\]But, if we have \(r<1\) we have a chance that it is convergent:
\[\begin{eqnarray} S = \sum_{i=0}^\infty a_i &=& 1 + r + r^2 + \cdots\\ rS &=&\quad\ \ r + r^2 + \cdots \end{eqnarray}\]Now we can substract \(S - rS\) and that way we get:
\[S-rS = 1\Rightarrow (1-r)S=1\Rightarrow S={1\over (1-r)}\] \[\hspace{25cm}\blacksquare\]If we want a more general Geometric Series:
\[\begin{eqnarray} a_0 &=& A \\ a_{i+1} &=& a_i\cdot r \end{eqnarray} \qquad\text{where }r\text{ is a real constant}\]we can see quickly that:
\[A + Ar + Ar^2 + \cdots = A(1+r+r^2+\cdots) = AS = {A \over (1-r)}\]Now the true interesting one:
The Harmonic series
We saw before that we checked whether the sequence “went to zero” before checking the series. We need this property, if we do not have it, the series cannot converge, this is why the geometric series, with ratio \(r\ge 1\) could not converge.
It seems intuitive that that would also go the other way, so to say, that if we do have the property, the series converges, but this is not the case, and what better way to prove this than a counterexample:
The harmonic series is simply defined as:
\[\lbrace a_i \rbrace_{i\in \mathbb{N}}\qquad\text{where } a_i = {1\over i}\]And, it would make sense for it to converge, right?2 To see that it does not, we will check if we can find a “smaller” series that does not converge either:
\[\begin{eqnarray} 1 + {1\over2} + {1\over3} + {1\over4} + {1\over5} +{1\over6}+{1\over7}+{1\over8}+{1\over9}+{1\over10}+ \cdots \ge\\ 1 + {1\over2} + {1\over4} + {1\over4} + {1\over8} +{1\over8}+{1\over8}+{1\over8}+{1\over16}+{1\over116}+ \cdots \end{eqnarray}\]We have substituted every element whose denominator is not a multiple of two by the closest multiple of two above it:
\[\begin{eqnarray} &1& + {1\over2} + {1\over4} + {1\over4} + {1\over8} +{1\over8}+{1\over8}+{1\over8}+{1\over16}+{1\over16}+ \cdots = \\ &=& (1) + \left({1\over2}\right) + \left({1\over4} + {1\over4}\right) + \left({1\over8} +{1\over8}+{1\over8}+{1\over8}\right) + \left({1\over16}+\cdots+{1\over16}\right)+\cdots \end{eqnarray}\]And now, each of the parenthesis sums to \(1\over2\), therefore, the above will equal:
\[1 + {1\over2} +{1\over2} +{1\over2} +\cdots = \infty\]Which means:
\[\begin{eqnarray} 1 + {1\over2} + {1\over3} + {1\over4} + {1\over5} +{1\over6}+{1\over7}+{1\over8}+ \cdots \ge 1 + {1\over2} +{1\over2} +{1\over2} +\cdots = \infty \end{eqnarray}\]Which trivially shows that the sum goes to infinity (because it is “greater than infinity”)
\[\hspace{25cm}\blacksquare\]