8. Week 7: Convergence of Series
≪ 7. Week 6: Infinite Finite Integrals and Finite Infinite Integrals and Infinite Infinite Integrals | Table of ContentsToday, our goal is to develop some (but not all) of the tools needed to study when infinite series converge. As a warm-up, let us remind ourselves of what series are defined as:
Warm-Up 1.
Suppose \(a_1, a_2, a_3, \ldots\) is a sequence of numbers. What is the mathematical definition of
\[ \sum _{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \ldots ? \]Note that it doesn’t make sense to just add infinitely many things together. Lurking in the background is a limiting statement: the more terms you add, the closer you should get to “the answer”. So, we define
\[ \sum _{n=1}^{\infty} a_n = \lim _{N\to\infty} \sum _{n=1}^{N} a_n = \lim _{N\to\infty} \left( a_1 + a_2 + \cdots + a_N \right). \]You may sometimes see \(\sum _{n=1}^{N}a_n\) abbreviated as \(S_N\).
From here, the story parallels that of improper integrals from last week: sometimes, these partial sums and these limits can actually be computed; however, these are few and far between. After that, there are some comparison theorems that help us identify which series converge and which ones don’t, even when we can’t compute them directly.
Computable Series
We begin with two classes of infinite series that are computable: the geometric and telescoping series.
Fact of Life 2. Geometric Series
A series of the form \(\sum _{n=0}^{\infty}r^n\) diverges if \(\left\lvert r \right\rvert \geq 1\) and converges to
\[ \sum _{n=0}^{\infty} r^n = \frac{1}{1-r} \]if \(\left\lvert r \right\rvert < 1\). In any case, the partial sums are given by the formula
\[\begin{align*} \sum _{n=0}^{N} r^n = \frac{1 - r^{N+1}}{1 - r}. \end{align*}\]This is a very important pair of formulae to remember! I don’t think there’s much else to say about it.
Another class of computable series are the telescoping series, where there is a lot of cancellation that makes the partial sums particularly simple. Let us observe the following example:
Example 3.
Compute the following series, or state that it diverges:
\[ \sum _{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right). \]If we write out the first several partial sums, a clear pattern begins to emerge:
\[\begin{align*} S_1 &= 1 - \frac{1}{2}, \\ S_2 &= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = 1 - \frac{1}{3}, \\ S_3 &= 1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +\frac{1}{3}-\frac{1}{4} = 1 - \frac{1}{4}, \\ &\vdots \\ S_N &= 1-\frac{1}{2} + \frac{1}{2}-\frac{1}{3} + \cdots + \frac{1}{N} - \frac{1}{N+1} = 1 - \frac{1}{N+1}. \end{align*}\]The “terms in the middle” always cancel each other out! Therefore, taking limits yields
\[ \sum _{n=1}^{\infty} \left( \frac{1}{n}-\frac{1}{n+1} \right) = \lim _{N\to\infty} S_N = \lim _{N\to\infty} \left( 1 - \frac{1}{N+1} \right) = 1. \]Uncomputable Series
Just like with our discussion on integrals, there are series whose values just cannot be computed, regardless of the amount of cleverness or raw ability one brings to the table. First, there’s a really easy way to tell if a series might diverge:
Theorem 4. Divergence Test
If \(\lim _{n\to\infty}a_n \neq 0\) or does not exist, then \(\sum _{n=1}^{\infty}a_n\) diverges.
If you’re adding infinitely many things, then those numbers have got to be getting smaller.
When \(\lim _{n\to\infty}a_n = 0\), it’s not necessarily true that the series will converge, and we have to appeal to an arsenal of more delicate tests.
Exercise 5.
Recall the integral comparison test, direct comparison test, and limit comparison test for the convergence of series.
Hopefully you did that before scrolling down. There are other tests that we’ll learn about in due time, but these are geometric in flavour and should remind you of some of the tests we learned for the convergence of improper integrals.
Theorem 6. Direct Comparison Test
Suppose \(a_1, a_2, a_3, \ldots\) and \(b_1, b_2, b_3, \ldots\) are two sequences such that \(0 \leq a_n \leq b_n\) for all \(n\). If \(\sum _{n=1}^{\infty} a_n\) diverges, then \(\sum _{n=1}^{\infty} b_n\) diverges too. If \(\sum _{n=1}^{\infty} b_n\) converges, then \(\sum _{n=1}^{\infty} a_n\) converges too.
Theorem 7. Limit Comparison Test
Suppose \(a_1, a_2, a_3, \ldots\) and \(b_1, b_2, b_3, \ldots\) are two sequences of nonnegative numbers such that \(0 < \lim _{n\to\infty} \frac{a_n}{b_n} < \infty\). Then, \(\sum _{n=1}^{\infty} a_n\) converges if and only if \(\sum _{n=1}^{\infty} b_n\) converges.
I should note here that the positivity of the limit is very important… what could happen if the limit is \(0\)?
Theorem 8. Integral Comparison Test
Suppose \(a_n = f(n)\) for a nonnegative nonincreasing function \(f(x)\) defined on \([1, \infty)\). Then \(\sum _{n=1}^{\infty} f(n)\) converges if and only if \(\int _{1}^{\infty} f(x) dx\) converges.
Draw a picture involving Riemann sums! The condition that \(f\) is nonincreasing is very important, and hopefully you can see where the visual reasoning fails.
This last comparison theorem, when coupled with the \(p\)-test for integrals, gives us a \(p\)-test for series as well. This one only concerns the “fatness of the tails”.
Corollary 9. Series \(p\)-test
\(\sum _{n=1}^{\infty } \frac{1}{n^p}\) converges if \(p > 1\) and diverges if \(p \leq 1\).
In practise, I don’t think the integral comparison test is very easy to apply, though in rare cases it’s the only one of the three that helps.
When looking at a series that may or may not converge, you should first apply the divergence test. If it tells you the series diverges, yay; otherwise, you’ll have to try something else. Usually, the summands will be fractions, and you should try to identify the “dominant term”. From there, you can apply some of the comparison tests to simmer the issue of convergence down to a series (or integral) involving only the dominant term. (This is a very fat usually…)
Exercise 10.
Determine whether or not the following series converge:
- \(\sum _{n=1}^{\infty} \frac{n}{3^n}\).
- \(\sum _{n=2}^{\infty} \frac{1}{n \left( \ln n \right)^2}\).
- \(\sum _{n=1}^{\infty} \frac{(5n+7) ^{1.5}}{(8n+19) ^{2.5}}\).
- \(\sum _{n=1}^{\infty} \frac{4n + n \sin n}{3n+2}.\)