Lecture 26 Convergence Tests Part 1

Text References: Course notes pp. 114-129 & Rogawski 10.2-10.3

26.1 Recap

Last time, we discussed sequences, sequences of partial sums, and series. We also started a discussion around convergence and divergence of infinite series.

Exercise 26.1 Consider the sequence \(\{a_k\}=3x^k\).

  1. List \(a_0,\ldots, a_3\).
  2. Calculate \(s_0,\ldots, s_3\).

Solution. For \(\{a_k\}=3x^k\), we have

  1. \(a_0=3x^0=3, \quad a_1 =3x^1=3x, \quad a_2=3x^2, \quad a_3=3x^3\)
  2. \[\begin{align*} s_0 &= 3 \\ s_1 &= 3+3x \\ s_2 &= 3+3x +3x^2\\ s_3 &= 3+3x +3x^2 +3x^3 \end{align*}\]

26.2 Learning Objectives

  • Determine whether a given geometric series converges or diverges.
  • Apply the \(n\)th Term Test to determine whether a series diverges.

26.3 Geometric Series

A series that is commonly encountered is the geometric series:

Definition 26.1 A geometric series has the form \[\sum_{k=0}^\infty ar^k = a+ ar+ ar^2+ar^3+\cdots\] The value of \(r\) is called the common ratio.

The key result for geometric series is:

Theorem 26.1 Let \(\sum_{k=0}^\infty ar^k\) be a geometric series.

  • If \(|r|<1\), then \(\displaystyle \sum_{k=0}^\infty ar^k =\dfrac{a}{1-r}\)
  • If \(|r|\geq 1\), then the geometric series diverges

Exercise 26.2 Determine whether the following geometric series converge or diverge. If they converge, give their value.

  1. \(\displaystyle \sum_{k=0}^\infty (-7)^k\)
  2. \(\displaystyle \sum_{k=0}^\infty \left ( \frac{1}{7} \right)^k\)

Solution. We have

  1. \(|r|=|-7|=7>1\) so this series diverges.
  2. \(|r|=\frac{1}{7}<1\) so this series converges to \(\dfrac{1}{1-\frac{1}{7}}=\dfrac{7}{6}\)

Exercise 26.3 Determine whether the series \(\displaystyle \sum_{k=4}^\infty 5 \left (\frac{1}{2} \right)^{k+2}\) converges and, if so, give its value.

Sometimes, a geometric series might not appear in the tidiest form at first glance. In cases like these, we might need to use re-indexing or other tricks to help us out.

Solution. Let’s start by writing out the first few terms in the sequence: \[\sum_{k=4}^\infty 5 \left (\frac{1}{2} \right)^{k+2} = 5\left ( \frac{1}{2} \right)^6 + 5\left ( \frac{1}{2} \right)^7 + 5\left ( \frac{1}{2} \right)^8+\cdots \]

Factoring out a \(\left ( \dfrac{1}{2}\right)^6\), we get \[\begin{align*} \sum_{k=4}^\infty 5 \left (\frac{1}{2} \right)^{k+2} &= 5\left ( \frac{1}{2} \right)^6 + 5\left ( \frac{1}{2} \right)^7 + 5\left ( \frac{1}{2} \right)^8+\cdots \\ & = \left ( \frac{1}{2} \right)^6 \left(5\left ( \frac{1}{2}\right )^0+5\left ( \frac{1}{2}\right )^1+5\left ( \frac{1}{2}\right )^2 +\cdots \right)\\ &= \sum_{k=0}^\infty \frac{5}{64} \left ( \frac{1}{2}\right )^k \end{align*}\]

We now have a tidier expression with \(a=\dfrac{5}{64}\) and \(r=\dfrac{1}{2}\). Since \(|r|<1\), this sequence will converge to \(\dfrac{\frac{5}{64}}{1-\frac{1}{2}}=\dfrac{5}{32}\).

26.4 The \(n\)th Term Test

Another test we can use to determine whether a series diverges is the \(n\)th Term Test.

Theorem 26.2 If \(\displaystyle \lim_{k\to\infty}a_k\neq 0\), then \(\displaystyle \sum a_k\) diverges.

Exercise 26.4 Show that \(\displaystyle \sum_{k=1}^\infty \frac{k}{4k+1}\) diverges.

Solution. Using the \(n\)th Term Test, we have \[\lim_{k\to\infty}a_k = \lim_{k\to\infty}\frac{k}{4k+1} = \lim_{k\to\infty} \frac{1}{4+\frac{1}{k}}=\frac{1}{4} \neq 0\] Therefore, by the \(n\)th Term Test, the series diverges.

Exercise 26.5 Show that \(\displaystyle \sum_{k=1}^\infty (-1)^{k-1}\frac{k}{k+1}=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}=\frac{4}{5}+\cdots\) diverges.

Exercise 26.6 Using the \(n\)th Term Test, we have \[\lim_{k\to\infty}a_k = \lim_{k\to\infty}(-1)^{k-1}\frac{k}{k+1}\] The odd terms, \(k=2n+1\), tend to \(1\); however, the even terms, \(k=2n\), tend to \(-1\). This means that \(\displaystyle \lim_{k\to\infty}a_k\) does not exist, and therefore the series diverges by the \(n\)th Term Test.

A very important observation about the \(n\)th Term Test is that the result is not and “if and only if” statement. In other words, we know what if the limit is not equal to zero, then the series diverges; however, we don’t know what happens if the limit is zero.