Comparison Tests for Convergence (College Board AP® Calculus BC): Revision Note

Useful series to use for comparisons are

• the geometric series $\sum _{n=0}^{\infty }a{r}^n$ with $a>0,r>0$, which converges for $0<r<1$ and diverges for $r\ge 1$

• the p-series $\sum _{n=1}^{\infty }\frac{1}{n^p}$, which converges for $p>1$ and diverges for $p\le 1$

Be sure to justify the use of this test by stating or showing that all the terms are non-negative.

Use the comparison test to determine whether each of the following series converges or diverges.

(a) $\sum _{n=1}^{\infty }\frac{1}{n^3+4}=\frac{1}{5}+\frac{1}{12}+\frac{1}{31}+...$

(b) $\sum _{n=1}^{\infty }\frac{n+3}{n^2+1}=2+1+\frac{3}{5}+...$

Answer:

(a)

The comparison here will be with $\sum _{n=1}^{\infty }\frac{1}{n^3}$, which is a convergent p-series

Note that $n^3+4>n^3$ for $n\ge 1$ so $\frac{1}{n^3+4}<\frac{1}{n^3}$

For all $n\ge 1$, $0<\frac{1}{n^3+4}<\frac{1}{n^3}$

$\sum _{n=1}^{\infty }\frac{1}{n^3}$ is a convergent p-series with $p=3>1$

By the comparison test, $\sum _{n=1}^{\infty }\frac{1}{n^3+4}$ converges

(b)

The comparison here is a lot less obvious!

Thinking that $\frac{n+3}{n^2+1}\approx \frac{n}{n^2}=\frac{1}{n}$ suggests using the divergent harmonic series $\sum _{n=1}^{\infty }\frac{1}{n}$

Method 1

There's a bit of algebra required to get an inequality in the desired form

For all $n\ge 1$,

$\begin{array}{rcl}\frac{n+3}{n^2+1}& =& \frac{n+1}{n^2+1}+\frac{2}{n^2+1}\\ & >& \frac{n+1}{n^2+1}\\ & =& \frac{n+1}{n^2+1}·\frac{n}{n}\\ & =& \frac{n^2+n}{n^3+n}\\ & \ge & \frac{n^2+1}{n^3+n}\\ & =& \frac{n^2+1}{n\left(n^2+1\right)}\\ & =& \frac{1}{n}\end{array}$

So $\begin{array}{rcl}& & \frac{n+3}{n^2+1}\end{array}>\frac{1}{n}>0$

And $\sum _{n=1}^{\infty }\frac{1}{n}$ is the harmonic series, which diverges

By the comparison test, $\sum _{n=1}^{\infty }\begin{array}{rcl}& & \frac{n+3}{n^2+1}\end{array}$ diverges

Method 2

Start with the result you want and ask if it is true, i.e. is $\frac{1}{n}<\frac{n+3}{n^2+1}$ true, where $n\ge 1$?

Assume it is true and rearrange it to get something that is actually true (this should be done in rough)

Be careful not to multiply both sides of an inequality by something negative (remember that $n\ge 1$ which is positive)

$\begin{array}{rcl}\frac{1}{n}& <& \frac{n+3}{n^2+1}\\ \left(n^2+1\right)& <& n\left(n+3\right)\\ n^2+1& <& n^2+3n\\ 1& <& 3n\\ \frac{1}{3}& <& n\end{array}$

The last line is actually true, since $n\ge 1$

Reverse the steps (starting with the line that is actually true) to get the algebraic proof you require

$\begin{array}{rcl}\frac{1}{3}& <& n\\ 1& <& 3n\\ n^2+1& <& n^2+3n\\ \left(n^2+1\right)& <& n\left(n+1\right)\\ 0& <& \frac{1}{n}<\frac{n+1}{n^2+1}\end{array}$

And $\sum _{n=1}^{\infty }\frac{1}{n}$ is the harmonic series, which diverges

By the comparison test, $\sum _{n=1}^{\infty }\begin{array}{rcl}& & \frac{n+3}{n^2+1}\end{array}$ diverges

For series expressed by more complicated rational functions, the limit comparison test can be a lot quicker and simpler to use than the comparison test as shown in the following worked example.

Use the limit comparison test to determine whether the series $\sum _{n=1}^{\infty }\frac{n+3}{n^2+1}=2+1+\frac{3}{5}+...$ converges or diverges.

Answer:

Considering only the highest powers in the numerator and denominator of $\frac{n+3}{n^2+1}$, you get $\frac{n}{n^2}=\frac{1}{n}$

This suggests using the divergent harmonic series $\sum _{n=1}^{\infty }\frac{1}{n}$ as the comparison series in the limit comparison test

$\frac{n+3}{n^2+1}>0$ and $\frac{1}{n}>0$ for $n\ge 1$

First set up and rewrite the quotient (it doesn't matter which series term goes on the top, and which goes on the bottom)

$\begin{array}{rcl}\frac{\left({\displaystyle \frac{n+3}{n^2+1}}\right)}{\left({\displaystyle \frac{1}{n}}\right)}& =& \frac{n+3}{n^2+1}·\frac{n}{1}\\ & =& \frac{n^2+3n}{n^2+1}\end{array}$

Then prepare the resulting expression for taking limits (by dividing the terms on top and bottom by the highest power of $n$)

$\begin{array}{rcl}& =& \frac{n^2+3n}{n^2+1}·\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n^2$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n^2$}\right.}}\\ & =& \frac{1+{\displaystyle \frac{3}{n}}}{1+{\displaystyle \frac{1}{n^2}}}\end{array}$

Now take the limit

$\begin{array}{rcl}\underset{n\to \infty }{\lim }\frac{\left({\displaystyle \frac{n+3}{n^2+1}}\right)}{\left({\displaystyle \frac{1}{n}}\right)}& =& \underset{n\to \infty }{\lim }\frac{1+{\displaystyle \frac{3}{n}}}{1+{\displaystyle \frac{1}{n^2}}}=\frac{1+0}{1+0}=1\end{array}$

That limit is positive and finite, so either both series converge or both diverge

$\sum _{n=1}^{\infty }\frac{1}{n}$ is the harmonic series, which diverges

By the limit comparison test, $\sum _{n=1}^{\infty }\begin{array}{rcl}& & \frac{n+3}{n^2+1}\end{array}$ diverges