Lecture 28 Convergence Tests Part 3

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

28.1 Recap

Last time, we stated convergence results for p-series and stated the Integral Test.

Exercise 28.1 Determine whether \(\displaystyle \sum_{k=0}^\infty e^{-k}\) converges or diverges.

Solution. The function \(f(x)=e^{-x}\) is positive, continuous, and decreasing for \(x\geq 0\). We can therefore apply the Integral Test:

\[\begin{align*} \int_0^\infty e^{-x} &= \lim_{t\to\infty} \int_0^t e^{-x} dx \\ &= \lim_{t\to\infty} \left. -e^{-x} \right|_{0}^t\\ &= \lim_{t\to\infty} 1-e^{-t} = 1 \end{align*}\]

Since the integral converges, the series does as well by the Integral Test.

28.2 Learning Objectives

Apply the Comparison Test appropriately to determine the convergence of a series.
Apply the Limit Comparison Test appropriately to determine the convergence of a series.

28.3 The Comparison Test

The next set of tests we’re going to learn about involve comparing a series with other series whose behaviour we already know; typically, we compare to p-series or geometric series.

Let’s start with the Comparison Test:

Theorem 28.1 Let \(\displaystyle \sum a_k\) be a series with all positive terms.

If there is a series \(\displaystyle \sum b_k\) such that \(a_k\leq b_k\) for all \(k\) and \(\displaystyle \sum b_k\) converges, then \(\displaystyle \sum a_k\) also converges.
If there is a series \(\displaystyle \sum b_k\) such that \(a_k\geq b_k\) for all \(k\) and \(\displaystyle \sum b_k\) diverges, then \(\displaystyle \sum a_k\) also diverges

Exercise 28.2 Determine whether \(\displaystyle \sum_{k=1}^\infty \frac{5}{2k^2+4k+3}\) converges or diverges.

Solution. For large values of \(k\), the dominant term in the denominator is \(2k^2\). Since \(\dfrac{5}{2k^2+4k+3}<\dfrac{1}{k^2}\) for all \(k\), we have \(\displaystyle \sum_{k=1}^\infty \frac{5}{2k^2+4k+3}< \displaystyle \sum_{k=1}^\infty \frac{5}{2k^2}\). Referring to the statement of the test, we have \(\displaystyle \sum a_k\) on the left and \(\displaystyle \sum b_k\) on the right.

Now we need to determine whether \(\displaystyle \sum_{k=1}^\infty \frac{5}{2k^2}\) converges or diverges. We can rewrite it as \(\displaystyle \frac{5}{2} \sum_{k=1}^\infty \frac{1}{2k^2}\), which we recognize as a p-series with \(p=2>1\) and therefore converges. Therefore, by the comparison test, \(\displaystyle \sum_{k=1}^\infty \frac{5}{2k^2+4k+3}\) converges.

Exercise 28.3 Determine whether \(\displaystyle \sum_{k=1}^\infty \frac{1}{\sqrt{k}3^k}\) converges or diverges.

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

Now we need to determine whether \(\displaystyle \sum_{k=1}^\infty \frac{1}{3^k}\) converges or diverges. We can rewrite it as \(\displaystyle \sum_{k=1}^\infty \left (\frac{1}{3}\right)^k\), which we recognize as a geometric series with \(r=1/3<1\) and therefore converges. Therefore, by the comparison test, \(\displaystyle \sum_{k=1}^\infty \frac{1}{\sqrt{k}3^k}\) converges.

28.4 The Limit Comparison Test

Theorem 28.2 If \(\displaystyle \lim_{k\to\infty}\frac{a_k}{b_k}=L\) where \(L\) is a nonzero constant, then \(\displaystyle \sum a_k\) and \(\displaystyle \sum b_k\) either both converge or both diverge.

The idea here is to take the limit of the ratio of the series we want to know about and another series whose behaviour we already know.

Exercise 28.4 Determine whether the series \(\displaystyle \sum_{k=1}^\infty \frac{1}{2^k-1}\) converges or diverges.

Solution. Let’s set \(a_k=\dfrac{1}{2^k-1}\) and \(b_k=\dfrac{1}{2^k}\). We know that \(\displaystyle \sum b_k\) is a geometric series with \(r=1/2<1\) and therefore converges.

We have

\[\begin{align*} \lim_{k\to\infty}\frac{a_k}{b_k}&= \lim_{k\to\infty} \dfrac{\frac{1}{2^k-1}}{\frac{1}{2^k}}\\ &= \lim_{k\to\infty}\dfrac{2^k}{2^k-1}\\ &= \lim_{k\to\infty}\dfrac{1}{1-1/2^k} =1 \end{align*}\]

Since the limit is a constant, the Limit Comparison Test tells us that the behaviour of \(\displaystyle \sum a_k\) is the same as the behaviour of \(\displaystyle \sum b_k\). Since \(\displaystyle \sum b_k\) converges, \(\displaystyle \sum a_k\) converges as well.

Exercise 28.5 Determine whether the series \(\displaystyle \sum_{k=2}^\infty \frac{1}{\sqrt{k^2+4}}\) converges or diverges.

Solution. Let’s set \(a_k=\frac{1}{\sqrt{k^2+4}}\) and \(b_k=\frac{1}{k}\). We know that \(\displaystyle \sum b_k\) is the harmonic series, which diverges.

We have

\[\begin{align*} \lim_{k\to\infty}\frac{a_k}{b_k}&= \lim_{k\to\infty} \dfrac{\frac{1}{\sqrt{k^2+4}}}{\frac{1}{k}}\\ &= \lim_{k\to\infty}\dfrac{k}{\sqrt{k^2+4}}\\ &= \lim_{k\to\infty}\dfrac{1}{\sqrt{1+4/k^2}} =1 \end{align*}\]

Since the limit is a constant, the Limit Comparison Test tells us that the behaviour of \(\displaystyle \sum a_k\) is the same as the behaviour of \(\displaystyle \sum b_k\). Since \(\displaystyle \sum b_k\) diverges, \(\displaystyle \sum a_k\) diverges as well.