Class 22 - Term test for divergence and Limit comparison test (10/26/2017, Thursday)

Exam 2

Your exam 2 is on November 13, and we have less than a month left.

Team HW

Your Team HW 4 is due 10/31 Tuesday.

Webwork

• 9.1 due 10/26 Thu
• 9.2 due 10/30 Mon

Term test for divergent

We have seen the following test in Quiz 4:

Term test for divergence. The series $\sum_{n=1}^{\infty}a_{n}$ diverges if $\lim_{n \rightarrow \infty} a_{n} \neq 0$.

Warning. Remember the name. You cannot test any convergence using this test. For example, take $a_{n} = 1/n$. We have $\lim_{n \rightarrow \infty}a_{n} = \lim_{n \rightarrow \infty}1/n = 0$, but

$$\sum_{n=1}^{\infty}a_{n} = \sum_{n=1}^{\infty}\frac{1}{n} = \infty$$

because it is a $p$-series with $p = 1 \leq 1$.

Many people seem to be confused about this, every semester!

Limit comparison test

Limit comparison test for positive series. Let $(a_{n})_{n \geq 0}$ and $(b_{n})_{n \geq 0}$ be positive sequences (i.e., all the terms $a_{n}, b_{n} > 0$). Then

1. If $\lim_{n \rightarrow \infty} a_{n}/b_{n}$ converges to a nonzero number, then $\sum_{n=1}^{\infty}a_{n}$ converges precisely when $\sum_{n=1}^{\infty}b_{n}$ converges. (That is, they either converge together or diverge together.)
2. If $\lim_{n \rightarrow \infty} a_{n}/b_{n} = 0$, then the convergence of $\sum_{n=1}^{\infty}b_{n}$ implies the convergence of $\sum_{n=1}^{\infty}a_{n}$.
3. If $\lim_{n \rightarrow \infty} a_{n}/b_{n} = \infty$, then $\sum_{n=1}^{\infty}b_{n} = \infty$ implies $\sum_{n=1}^{\infty}a_{n} = \infty$.

Proof (optional reading). To prove 1, suppose that we have
$$\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = c \neq 0.$$ Since $a_{n}/b_{n} \geq 0$ for all $n$, we must have $c \geq 0$, so we necessarily have $c > 0$. Take any $\epsilon > 0$ such that $c > \epsilon$. Then for $n \geq N$ for large enough $N$, we have $a_{n}/b_{n} \in (c-\epsilon, c+\epsilon)$. In terms of inequalities, this means that
$$0 < c - \epsilon < a_{n}/b_{n} < c + \epsilon.$$ Hence, we have $(c - \epsilon)b_{n} < a_{n}$ and $a_{n} < (c + \epsilon)b_{n}$. Thus, by comparison test, the first inequality tells us that $\sum_{n=N}^{\infty}b_{n}$ converges if $\sum_{n=N}^{\infty}a_{n}$ does, while the second inequality tells us that $\sum_{n=N}^{\infty}a_{n}$ converges if $\sum_{n=N}^{\infty}b_{n}$ does. Finitely many previous terms do not matter when we test convergence, so this shows 1.

To prove 2, suppose that
$$\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = 0.$$ For $n \geq N$ for large enough $N$, we have $a_{n}/b_{n} \leq 1$, so $a_{n} \leq b_{n}$. Hence by comparison test, the series $\sum_{n=N}^{\infty}a_{n}$ converges when $\sum_{n=N}^{\infty}b_{n}$ converges.

To prove 3, suppose that
$$\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = \infty.$$ For $n \geq N$ for large enough $N$, we have $a_{n}/b_{n} \geq 1$, so $a_{n} \geq b_{n}$. Hence by comparison test, the series $\sum_{n=N}^{\infty}a_{n} = \infty$ when $\sum_{n=N}^{\infty}b_{n} = \infty$.

Q.E.D.

Exercise. Determine whether the following series converges:

$$\sum_{n=1}^{\infty}\frac{n^{5} + n^{4} + 220n + 30}{2n^{6} + 5}.$$