(S3) Introduction to Series#

In this lesson we are going to see:

  • the limit definition of series using partial sums.

  • the algebraic properties of convergent series.

  • how to write out the first few terms of the sequence of partial sums for a given series.

Review Videos#

Series#

(Infinite) Series

Consider the infinite sequence \(\displaystyle \left\{ a_n\right\}_{n=1}^{\infty}\). A series is an expression of the form:

\[ a_1+a_2+a_3+a_4+ \cdots + a_n+ \cdots \]

which can also be written as

\[ \sum_{n=1}^{\infty} a_n \quad \text{or} \quad \sum a_n \]

Intuitive Definition#

Essentially, we are trying to add all of the terms of an infinite sequence (in a specific order). As we’re going to see, this cannot always be done:

  • If the sum exists we say the series is convergent.

  • If the sum does not exist, we say the series is divergent.

How do we add infinitely many terms?#

Adding infinitely many terms is not like our usual \(+\) operation. Sometimes we can do it, sometimes we can’t. But as with anything involving infinity, the general strategy involves limits:

  1. Add a finite number of terms.

  2. Compute the limit as the number of terms we are adding approaches infinity.

Partial Sums#

We start by considering what happens when we add a finite number of terms from our sequence \(\{a_n\}\). The result is called a partial sum.

Partial Sums

Given a series \(\displaystyle \sum_{n=1}^{\infty} a_n\), we let \(s_n\) denote the \(n^{\text{th}}\) partial sum:

\[ s_n = \sum_{i=1}^{n} a_i = a_1+a_2+a_3+a_4+ \cdots + a_n \]

Sequence of Partial Sums#

So given a sequence \(\{a_1, a_2, a_3, \dots, a_n, \dots \}\) we can form a brand new sequence\( \{s_1, s_2, s_3, \dots, s_n, \dots \}\), called the sequence of partial sums, where:

\[\begin{split} s_1 & = a_1 \\[5pt] s_2 & = a_1+a_2 \\[5pt] s_3 & = a_1+a_2 +a_3 \\[5pt] s_4 & = a_1+a_2 +a_3 +a_4\\[5pt] \end{split}\]

Example 1#

Write out the first 4 terms of the sequence of partial sums, for the series:

\[ \dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\cdots + \dfrac{1}{3^n}+\cdots \]
Solution (Click to see the steps.)

We start by writing out the original sequence \(\{a_n\}\).

\[ \{a_n\}_{n=1}^{\infty} = \left\{ \dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{27},\dfrac{1}{81},\dots , \dfrac{1}{3^n},\dots \right\} \]

From here we add progressively more and more of the terms together to get our partial sums:

\[\begin{split} s_1 & = a_1 & = \dfrac{1}{3} & = \dfrac{1}{3}\\[5pt] s_2 & = a_1 +a_2 & = \dfrac{1}{3} +\dfrac{1}{9} & = \dfrac{4}{9} \\[5pt] s_3 & = a_1 +a_2 +a_3 & = \dfrac{1}{3} +\dfrac{1}{9}+\dfrac{1}{27} & = \dfrac{13}{27}\\[5pt] s_4 & = a_1 +a_2 +a_3 +a_4 \quad & = \dfrac{1}{3} +\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81} \quad & = \dfrac{40}{81}\\[5pt] \end{split}\]

The sequence of partial sums is therefore:

\[ \{s_n\}_{n=1}^{\infty} = \left\{ \dfrac{1}{3}, \dfrac{4}{9}, \dfrac{13}{27}, \dfrac{40}{81}, \dots \right\} \]

Example 2#

Write out the first 5 terms of the sequence of partial sums, for the series:

\[ \sum_{n=1}^{\infty} \dfrac{1}{n} \]
Solution (Click to see the steps.)

We start by writing out the original sequence \(\{a_n\}\).

\[ \{a_n\}_{n=1}^{\infty} = \left\{ 1, \dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dots , \dfrac{1}{n},\dots \right\} \]

From here we add progressively more and more of the terms together to get our partial sums:

\[\begin{split} s_1 & = a_1 & = 1 & = 1\\[5pt] s_2 & = a_1 +a_2 & = 1 +\dfrac{1}{2} & = \dfrac{3}{2} \\[5pt] s_3 & = a_1 +a_2 +a_3 & = 1 +\dfrac{1}{2} +\dfrac{1}{3} & = \dfrac{11}{6}\\[5pt] s_4 & = a_1 +a_2 +a_3 +a_4 \quad & = 1 +\dfrac{1}{2} +\dfrac{1}{3}+\dfrac{1}{4} \quad & = \dfrac{25}{12}\\[5pt] s_5 & = a_1 +a_2 +a_3 +a_4 +a_5 \quad & = 1 +\dfrac{1}{2} +\dfrac{1}{3}+\dfrac{1}{4} +\dfrac{1}{5}\quad & = \dfrac{137}{60}\\[5pt] \end{split}\]

The sequence of partial sums is therefore:

\[ \{s_n\}_{n=1}^{\infty} = \left\{ 1, \dfrac{3}{2}, \dfrac{11}{6}, \dfrac{25}{12}, \dfrac{137}{60}, \dots \right\} \]

Series Convergence#

Convegence of a Series

A series \(\sum a_n\) is said to be convergent if the sequence \(\{s_n\}\) of partial sums converges to some number \(s\). This number \(s\) is called the sum of the series. And we write:

\[ \sum_{n=1}^{\infty} a_n = \lim_{n\to \infty} \sum_{i=1}^n a_i= \lim_{n\to \infty} s_n = s \]

On the other hand, a series \(\sum a_n\) is said to be divergent if the sequence \(\{s_n\}\) of partial sums diverges.

A few comments:

  1. It is usually difficult to determine a nice, simple formula for the general term \(s_n\) of the sequence of partial sums.

  2. Even if we know a series converges, the exact sum is usually difficult to determine.

Example 3#

Show that the following series is convergent and find its sum.

\[\sum_{n=1}^{\infty} \dfrac{1}{n(n+1)}\]

(Click to see the steps.)

We start by performing a partial fraction decomposition of this general term (it has two distinct linear terms).

\[ \dfrac{1}{n(n+1)} = \dfrac{1}{n}-\dfrac{1}{n+1} \]

(try working the details of this for practice!)

We want to find the sum of this series, so this means we need to find the general term \(s_n\) of our sequence of partial sums. To help us find this general term we write out the first few terms of the sequence, and see if we can identify the pattern:

\[\begin{split} s_1 & = \dfrac{1}{1}-\dfrac{1}{2} \\[5pt] s_2 & = \left(\dfrac{1}{1}-\dfrac{1}{2}\right) + \left(\dfrac{1}{2}-\dfrac{1}{3}\right) \\[5pt] s_3 & = \left(\dfrac{1}{1}-\dfrac{1}{2}\right) + \left(\dfrac{1}{2}-\dfrac{1}{3}\right) + \left(\dfrac{1}{3}-\dfrac{1}{4}\right) \\[5pt] s_4 & = \left(\dfrac{1}{1}-\dfrac{1}{2}\right) + \left(\dfrac{1}{2}-\dfrac{1}{3}\right)+ \left(\dfrac{1}{3}-\dfrac{1}{4}\right)+ \left(\dfrac{1}{4}-\dfrac{1}{5}\right) \\[5pt] \end{split}\]

But notice that for each of these partial sums, all of the fractions cancel except for the first and the last (this is why this type of series is sometimes referred to as a telescoping series). Simplifying we get:

\[\begin{split} s_1 & = 1-\dfrac{1}{2} \\[5pt] s_2 & = 1-\dfrac{1}{3} \\[5pt] s_3 & = 1-\dfrac{1}{4} \\[5pt] s_4 & = 1-\dfrac{1}{5} \\[5pt] \end{split}\]

And from here we can identify what the general term \(s_n\) would be:

\[ s_n = 1-\dfrac{1}{n+1} \]

To determine the sum of the series, we then calculate the limit of our sequence of partial sums:

\[ \lim_{n\to\infty}s_n = \lim_{n\to\infty} \left(1-\dfrac{1}{n+1}\right) = 1-0=1 \]

Since the sequence of partial sums converges to 1, we say the series converges and has sum equal to 1.

\[ \sum_{n=1}^{\infty} \dfrac{1}{n(n+1)} =1 \]

Algebraic Properties#

Convergent series behave very nicely with addition, subtraction, and constant multiplication.

Algebraic Properties of Convergent Series

If \(\sum a_n\) and \(\sum b_n\) are convergent series, then so are the series:

\[\begin{split} \sum_{n=1}^{\infty} c a_n & = c \cdot \sum_{n=1}^{\infty} a_n \\[10pt] \sum_{n=1}^{\infty} \big(a_n+b_n \big) & = \sum_{n=1}^{\infty} a_n +\sum_{n=1}^{\infty} b_n \\[10pt] \sum_{n=1}^{\infty} \big(a_n-b_n \big) & = \sum_{n=1}^{\infty} a_n -\sum_{n=1}^{\infty} b_n \\[10pt]\end{split}\]

Warning

There are no rules like this for multiplication and division.

Example 4#

Determine the sum of the series:

\[\sum_{n=1}^{\infty} \left(\dfrac{5}{3^n}-\dfrac{1}{2^n}\right)\]

Given that we know:

\[ \sum_{n=1}^{\infty} \dfrac{1}{3^n} = \tfrac{1}{2} \qquad \text{and} \qquad \sum_{n=1}^{\infty} \dfrac{1}{2^n} = 1 \]
Solution (Click to see the steps.)

Applying the algebraic properties of convergent series, we can break our original series up:

\[\begin{split} \sum_{n=1}^{\infty} \left(\dfrac{5}{3^n}-\dfrac{1}{2^n}\right) & = \sum_{n=1}^{\infty} \dfrac{5}{3^n}-\sum_{n=1}^{\infty} \dfrac{1}{2^n} \\[10pt] & = 5 \cdot \sum_{n=1}^{\infty} \dfrac{1}{3^n}-\sum_{n=1}^{\infty} \dfrac{1}{2^n} \\[10pt] & = 5 \cdot \tfrac{1}{2}-1 \\[10pt] & = \tfrac{3}{2}\\[10pt] \end{split}\]

The reason why we’re able to do these steps here is because we know that the series \(\sum \tfrac{1}{3^n}\) and \(\sum \tfrac{1}{2^n}\) are both convergent. And then we also happen to know their sums, so we fill this in as our last step.