Lecture 27 Convergence Tests Part 2

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

27.1 Recap

Last time, we stated convergence results for geometric series and stated the \(n\)th Term Test.

Exercise 27.1 Show that \(\displaystyle \sum_{k=1}^\infty \frac{k^2}{5k^2+4}\) diverges.

Solution. We have \[\lim_{k\to\infty}a_k = \lim_{k\to\infty}\frac{k^2}{5k^2+4}=\lim_{k\to\infty}\frac{1}{5+4/k^2}=\frac{1}{5}\neq 0\]

Therefore, the series diverges by the \(n\)th Term Test.

27.2 Learning Objectives

  • Apply the Integral Test appropriately to determine the convergence of a series.
  • Apply the p-Series Test appropriately to determine the convergence of a series.

27.3 The Integral Test

The next test we’ll discuss is the Integral Test

Theorem 27.1 Let \(f\) be a continuous, positive, and decreasing function on \((k_0,\infty)\) with \(f(k)=a_k\) for all \(k\geq k_0\). Then, \(\displaystyle \sum_{k=k_0}^\infty a_k\) converges if and only if \(\displaystyle \int_{k_0}^\infty f(x)dx\) converges. In other words:

  • If \(\displaystyle \int_{k_0}^\infty f(x)dx\) converges, then \(\displaystyle \sum_{k=k_0}^\infty a_k\) converges
  • If \(\displaystyle \int_{k_0}^\infty f(x)dx\) diverges, then \(\displaystyle \sum_{k=k_0}^\infty a_k\) diverges

One important thing to note about the Integral Test is that it is able to tell us whether the series converges or diverges, but it doesn’t tell us anything about the value to which the series converges. The value of the improper integral is therefore less important for this test than whether the integral converges.

Exercise 27.2 Determine whether the series \(\displaystyle \sum_{k=1}^\infty \frac{1}{k^2}\) converges or diverges.

Solution. (Note that this is a case where the \(n\)th Term Test gives a limit of zero and is therefore inconclusive.)

The function \(f(x)=\frac{1}{x^2}\) is positive, continuous, and decreasing for \(x>0\). We can therefore apply the Integral Test:

\[\begin{align*} \int_1^\infty \dfrac{1}{x^2} &= \lim_{t\to\infty} \int_1^t \dfrac{1}{x^2} dx \\ &= \lim_{t\to\infty} \left. -\dfrac{1}{x} \right|_{1}^t\\ &= \lim_{t\to\infty} 1-\dfrac{1}{t} = 1 \end{align*}\]

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

Exercise 27.3 Determine whether the series \(\displaystyle \sum_{k=1}^\infty \frac{\ln(k)}{k}\) converges or diverges.

Solution. The function \(f(x)=\frac{\ln(x)}{x}\) is positive and continuous for \(x>1\). It is also decreasing when \(x>e\). We can therefore apply the Integral Test:

\[\begin{align*} \int_1^\infty \dfrac{\ln(x)}{x} &= \lim_{t\to\infty} \int_1^t \dfrac{\ln(x)}{x} dx \\ &= \lim_{t\to\infty} \left. \dfrac{(\ln(x))^2}{x} \right|_{1}^t\\ &= \lim_{t\to\infty} \dfrac{\ln(t)^2}{2} = \infty \end{align*}\]

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

27.4 The p-series Test

The p-series is a family of series of the form \(\displaystyle \sum \frac{1}{k^p}\) and has its corresponding test:

Theorem 27.2 The series \(\displaystyle \sum \frac{1}{k^p}\) converges if \(p>1\) and diverges if \(p\leq 1\).

Exercise 27.4 Determine whether the series:

  1. \(\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^3}\)
  2. \(\displaystyle \sum_{k=1}^\infty \dfrac{1}{{k}^{1/7}}\)
  3. \(\displaystyle \sum_{k=1}^\infty \dfrac{1}{k}\) This series is known as the harmonic series.

converge or diverge.

Solution. We recognize all of these as p-series, so we can apply the p-series Test:

  1. We have \(p>1\) so this series converges.
  2. We have \(p\leq 1\) so this series diverges.
  3. We have \(p\leq 1\) so this series diverges.