Calculus II Lesson 20: Convergence Tests

1. Presentations
2. Questions?
3. Divergence Test
   1. Converse?
4. Integral Test
   1. Exercises
   2. p-series
   3. Comparison
5. Alternating
   1. Alternating Harmonic Series
   2. Conditional Convergence
6. Comparison Test
7. Ratio
8. Upcoming

Presentations

Questions?

• Last week’s homework
• This week’s homework
• MyOpenMath?
• Previous lessons:
  • Arc length / areas in polar
  • Describing Sequences / closed forms
  • Zeno’s Paradox
  • $0.999\ldots$
  • Infinite series
  • Geometric series

Divergence Test

Geometric series are particularly nice, but there are other series as well. Let’s look at the constant series $\sum\limits_{n = 0}^{\infty} \frac{1}{2}$. It should be clear that this diverges: the partial sums are $\frac{1}{2}, 1, \frac{3}{2}, 2 \frac{5}{2}, \ldots$ which continue to get bigger and bigger. Similarly, the series $\sum\limits_{n = 0}^{\infty} \frac{1}{1000000}$ also diverges, even if it grows pretty slowly. Moreover there is a nice theorem that generalizes both of these:

Theorem: If $\lim\limits_{n \rightarrow \infty} a_n \neq 0$, then $\sum\limits_{n=0}^{\infty} a_n$ diverges.

That is: if the limit of a sequence either does not exist, or exists but is not 0, then the sum of that sequence diverges. This is known as the Divergence Test.

For example, the series $0 + \frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} + \ldots$ diverges, since the sequence can be written as $a_n = \frac{n}{n+1}$. This sequence approaches $1$, and so the series diverges.

Converse?

The Divergence Theorem tells you directly which series diverge. But does it tell you anything about which series must converge? Not necessarily. For example, let’s look at the “Harmonic Series” $\sum \frac{1}{n}$:

• $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \geq 2$, since $\frac{1}{3} + \frac{1}{4} \geq \frac{1}{2}$.
• $1 + \frac{1}{2} + \ldots + \frac{1}{8} \geq 2.5$, since $\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \geq \frac{1}{2}$
• $1 + \ldots + \frac{1}{16} \geq 3$, for similar reasons.

This series diverges! It goes off to infinity quite slowly, but it does, in fact, go off to infinity. How can we tell?

Integral Test

One way to tell is by graphing the function $f(x) = \frac{1}{x}$ and drawing in the “Riemann sum” approximations.

Notice that the rectangles are all above the graph, and, therefore, we can see:

\[\sum_{n = 1}^{\infty} \frac{1}{n} \geq \int_1^{\infty} \frac{1}{x} dx\]

Further, we know how to compute this integral: since the antiderivative of $\frac{1}{x}$ is $\ln(x)$, this integral can be computed as the limit as $b \rightarrow \infty$ of $\ln(b) - \ln(1)$, which is $\infty$. Therefore, the series must also diverge.

Similarly, let’s look at the series $\sum\limits_{n = 1}^{\infty} \frac{1}{n^2}$. We can similarly compare it to the integral of $f(x) = \frac{1}{x^2}$. In fact, there are two ways we can compare it:

and

Notice that this means that if

\[\int_1^\infty \frac{1}{x^2} dx\]

converges, then so does

\[\sum_{n=1}^{\infty} \frac{1}{n^2}\]

and if the integral diverges, then so does the series. And, again, we know how to compute the integral:

\[\begin{align} &\lim_{b \rightarrow \infty} \int_1^b \frac{1}{x^2} dx \\ &= \lim_{b \rightarrow \infty} (-\frac{1}{b} + 1) \\ &= 1 \end{align}\]

Since the integral converges, the series also converges. This is known as the Integral Test: suppose $(a_n)$ is a sequence of positive terms, and $f(x)$ is a function with the following properties:

• $f(x)$ is decreasing,
• $f(x)$ is continuous, and
• there is some $N$ such that for all $n \geq N$, $a_n = f(n)$

Then either \(\int_N^{\infty} f(x) dx \textrm{ and } \sum_{n = 1}^{\infty} a_n\) both converge or they both diverge. Note that this does not necessarily mean that they both converge to the same value: we can see that the integral of $\frac{1}{x^2}$ from $1$ to $\infty$ converged to $1$. But the series actually converges to $\frac{\pi^2}{6}$: this is known as the Basel problem.

Exercises

Use the integral test to determine if the following series converge or diverge:

1. $\sum\limits_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
2. $\sum\limits_{n=1}^{\infty} \frac{1}{n^{3/2}}$

p-series

In general:

• $\sum\limits_{n=1}^{\infty} \frac{1}{n^p}$ is called a $p$-series.
• A $p$-series converges if $p > 1$
• Diverges if $p \leq 1$.

Comparison

Notice:

\[\begin{align} 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + \ldots \\ \geq 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots \rightarrow \infty \end{align}\]

Since $\frac{1}{\sqrt{n}} \geq \frac{1}{n}$ for all $n \geq 1$.

Alternating

Definition: An alternating series is one in which the terms alternate between positive and negative.

Example: $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots$

\[= \sum_{n = 0}^{\infty} \frac{(-1)^{n}}{n+1}\]

Examples:

• $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots$ converges! ($\rightarrow \ln(2)$)
• $1 - 1 + 1 - 1 +1 \ldots$ diverges.
• $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots$…?

Does this converge?

This also converges! In fact this converges to $\frac{\pi}{4}$, giving us a formula to compute $\pi$! (This converges quite slowly, though...)

Theorem: Suppose either:

• $a_n = (-1)^n b_n$, or,
• $a_n = (-1)^{n+1} b_n$,

where $b_n \geq 0$, $\lim\limits_{n \rightarrow \infty} b_n = 0$, and $b_n$ is decreasing. Then $\sum\limits_{n=0}^{\infty} a_n$ converges.

In other words: any alternating series where the absolute values of the terms decrease and appraoch 0 converges!

Non-example: Does the series $1 - \frac{1}{4} + \frac{1}{2} - \frac{1}{16} + \frac{1}{4} - \frac{1}{64} + \frac{1}{8} - \ldots$ converge?

This test is inconclusive. $b_n = 1, \frac{1}{4}, \frac{1}{2}, \frac{1}{16}, \ldots$. Not decreasing!

Alternating Harmonic Series

Recall the Commutative law of addition:

• $x + y = y + x$. Right?
• $x_1 + x_2 + x_3 + \ldots + x_n = x_7 + x_{23} + \ldots + x_2$

In other words: if you add up a bunch of numbers, you can add them in whatever order you like. We already saw, for example, that the alternating harmonic series converges to $\ln(2)$: \(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \rightarrow \ln(2).\)

What happens when we rearrange?

\[\begin{align} &(1 - \frac{1}{2}) - \frac{1}{4} + (\frac{1}{3} - \frac{1}{6}) - \frac{1}{8} + (\frac{1}{5} - \frac{1}{10}) - \ldots \\ &=\frac{1}{2}(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots \end{align}\]

converges to $\frac{1}{2} \ln(2)$!

Conditional Convergence

• Alternating harmonic series converges to $\ln(2)$
• Rearranging some terms, it converges to $\frac{\ln(2)}{2}$
• This is because of conditional convergence!

Definitions:

• If a series $\sum \mid a_n \mid$ converges, we say it converges absolutely.
• If not, but $\sum a_n$ converges, we say it converges conditionally.

It turns out that series that converge conditionally do not respect the commutative law of addition! (This is crazy!)

In fact: if a series converges conditionally, then:

1. There is a way to rearrange it so that it diverges, and,
2. for each real number $r$, there is a way to rearrange the series so that it converges to $r$.

Comparison Test

Suppose $0 \leq a_n \leq b_n$. Then:

1. If $\sum b_n$ converges, then $\sum a_n$ converges.
2. If $\sum a_n$ diverges, then $\sum b_n$ diverges.

“If the bigger series converges, so does the smaller series. If the smaller series diverges, so does the bigger series.”

Example:

\[\sum_{n = 1}^{\infty} \frac{1}{n^2 + 1}\]

• Compare with $\sum\limits_{n = 1}^{\infty} \frac{1}{n^2}$: converges ($p$-series).
• Notice: $\frac{1}{n^2 + 1} \leq \frac{1}{n^2}$ for all $n$.
• (Algebra)
• The bigger series converges.
• So this also converges!

Example 2:

\[\sum_{n = 1}^{\infty} \frac{1}{2^n + n^2}\]

What should we compare this with?

Example 3:

\[1 + \frac{1}{3} + \frac{1}{5} + \ldots\]

What should we compare this with?

Ratio

The ratio test is one of the most important tests for convergence we can use. Given a series $\sum a_n$, look at the ratios:

\[\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| = L\]

• $L < 1$: absolute convergence
• $L > 1$: divergence
• $L = 1$: inconclusive

Example:

\[\sum_{n=0}^{\infty} \frac{1}{n!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \ldots\]

• $a_{n+1} = \frac{1}{(n+1)!}$
• $a_n = \frac{1}{n!}$
• $\frac{a_{n+1}}{a_n} = \frac{n!}{(n+1)!} = \frac{1}{n+1} \rightarrow 0$
• So this series converges absolutely!
• In fact: the series converges to $e$!

Exercise:

Determine if the following series converge:

1. $\sum_{n = 0}^{\infty} \frac{3^n}{n!}$
2. $\sum_{n = 0}^{\infty} \frac{3^n}{(-2)^{n+1} \cdot n}$

Upcoming

• Power series, Taylor series
• Exam 3: Monday, 4/27 or Thursday 4/30
• Final exam: May 7 9 AM - 11:30 AM.
• Math/CS Career Panel April 22 5 PM.