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
- 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
- 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.)
- If $\lim_{n \rightarrow \infty} a_{n}/b_{n} = 0$, then the convergence of $\sum_{n=1}^{\infty}b_{n}$ implies the convergence of $\sum_{n=1}^{\infty}a_{n}$.
- 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 $\sum_{n=N}^{\infty}b_{n} = \infty$.
\[\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = c \neq 0.\]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:
\[\sum_{n=1}^{\infty}\frac{n^{5} + n^{4} + 220n + 30}{2n^{6} + 5}.\]
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