Lecture 29 Convergence Tests Part 4

Text References: Course notes pp. 129-136 & Rogawski 10.3-10.5

29.1 Recap

Last time, we learned about the Comparison Test and the Limit Comparison Test.

Exercise 29.1 Determine whether \(\displaystyle \sum_{k=1}^\infty \frac{\ln(k)}{k}\) converges or diverges.

Solution. For large values of \(k\), the dominant term in the denominator is \(k\). Since \(\dfrac{\ln(k)}{k}>\dfrac{1}{k}\) for all \(k\geq 3\), we have \(\displaystyle \sum_{k=1}^\infty \frac{\ln(k)}{k}> \displaystyle \sum_{k=1}^\infty \frac{1}{k}\).

Now we need to determine whether \(\displaystyle \sum_{k=1}^\infty \frac{1}{k}\) converges or diverges. We recognize it as the harmonic series, which diverges. Therefore, by the Comparison Test, \(\displaystyle \sum_{k=1}^\infty \frac{\ln(k)}{k}\) diverges.

29.2 Learning Objectives

Apply the Alternating Series Test appropriately to determine the convergence of a series.
Use the Alternating Series Estimation Theorem to find the error of an approximation of the value of an alternating series.

29.3 The Alternating Series Test

Until now, the tests that we’ve developed only apply to series whose terms are positive. Now, we’re going to explore some tests for series whose terms are not necessarily positive.

Definition 29.1 An alternating series is a series whose terms alternate between positive and negative values. It has the general form \(\displaystyle \sum(-1)^k a_k\)

Given an alternating series, we can use the Alternating Series Test to determine its convergence:

Theorem 29.1 Consider the series \(\displaystyle \sum_{k=0}^\infty (-1)^k a_k=a_0-a_1+a_2-a_2+\cdots\) where \(a_k>0\) for every \(k\). If \(\displaystyle \lim_{k\to\infty}a_k=0\) and the sequence \(\{a_k\}\) is decreasing, then the series converges.

Exercise 29.2 Determine whether the alternating harmonic series \(\displaystyle \sum_{k=0}^\infty (-1)^k\frac{1}{k}\) converges or diverges.

Solution. We have that \(\displaystyle \lim_{k\to\infty}\frac{1}{k}=0\), and furthermore \(a_{k+1}<a_k\) since \(\dfrac{1}{k+1}<\dfrac{1}{k}\). Therefore, by the Alternating Series Test, the series converges.

Exercise 29.3 Determine whether the series \(\displaystyle \sum_{k=2}^\infty (-1)^k\sin\left (\frac{\pi}{k}\right )\) converges or diverges.

Solution. We have that \(\displaystyle \lim_{k\to\infty}\sin\left (\frac{\pi}{k}\right )=0\), and furthermore \(a_{k+1}<a_k\) since \(\sin\left (\frac{\pi}{k+1}\right )<\sin\left (\frac{\pi}{k}\right )\). Therefore, by the Alternating Series Test, the series converges.

29.4 The Alternating Series Estimation Theorem

A nice property of alternating series is that their value can be approximated without too much work. The Alternating Series Estimation Theorem (ASET) tells us how do proceed:

Theorem 29.2 Consider a convergent alternating series \(\displaystyle \sum (-1)^k a_k\). If we use the \(n\)th partial sum \(s_n\) as an estimate of the sum \(s\), then the error satisfies the inequality \[|s-s_n|\leq a_{n+1}\] In other words, the truncation error is less than the first omitted term in the partial sums.

Exercise 29.4 Estimate the sum of the series \(\displaystyle \sum_{k=0}^\infty \dfrac{(-1)^k}{k!}\) using the sixth partial sum. What is the accuracy of this estimate?

Solution. First, we need to verify whether the alternating series actually converges. We can use the Alternating Series Test:

\(\displaystyle \lim_{k\to\infty}\frac{1}{k!}=0\); and
\(a_{k+1}<a_k\) since \(\dfrac{1}{(k+1)!}<\dfrac{1}{k!}\)

Therefore the series converges.

We can now use the ASET to help us with the approximation. We have \[s\approx \frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!} \approx 0.366667\]

The seventh term is \(a_7\approx 0.00138889\). Therefore, \(|s-s_6|\leq 0.00138889\)

29.5 Absolute and Conditional Convergence

We’re now going to distinguish between two types of convergence.

Definition 29.2 A series \(\displaystyle \sum a_k\) is absolutely convergent if the series \(\displaystyle \sum |a_k|\) is convergent.

Definition 29.3 A series \(\displaystyle \sum a_k\) is conditionally convergent if the series is convergent, but \(\displaystyle \sum |a_k|\) is divergent

What we’re doing with these definition is distinguishing between series whose convergence does not change when we rearrange its terms (absolute convergence) and series whose convergence changes when we rearrange its terms (conditional convergence). You can find a more thorough discussion of this in the course notes.

We’ll also make note of a useful result:

Theorem 29.3 If a series \(\displaystyle \sum a_k\) is absolutely convergent, then it is also conditionally convergent.

Exercise 29.5 Determine whether the series \(\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2}\) is absolutely convergent.

Solution. We have \(\displaystyle \sum_{k=1}^\infty \left | \frac{(-1)^{k-1}}{k^2} \right | =\sum_{k=1}^\infty \frac{1}{k^2}\). This is a p-series with \(p=2\), and is therefore convergent. Hence, the series \(\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2}\) is absolutely convergent (and therefore also conditionally convergent).

Exercise 29.6 Determine whether the alternating harmonic series \(\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\) is absolutely convergent.

Solution. We have \(\displaystyle \sum_{k=1}^\infty \left | \frac{(-1)^{k-1}}{k} \right | =\sum_{k=1}^\infty \frac{1}{k}\). This is the harmonic series, which diverges. Hence, the series \(\displaystyle \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2}\) is not absolutely convergent. In fact, the alternating harmonic series is conditionally convergent.