The integral test gave us a starting point for determining the convergence or divergence of certain series. This was accomplished by comparing series to integrals.
In this section we begin comparing series to other series.
Example
Consider the pair of series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+1}.$
Comparing Series: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2+1}$
| Partial Sum $S_N$ | |||
| $N$ |
Example
Consider the pair of series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n-0.5}.$
Comparing Series: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$ and $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n-0.5}$
| Partial Sum $S_N$ | |||
| $N$ |
The Direct Comparison Test
Suppose there exists an integer $N$ such that $0 \leq a_n \leq b_n$ for all $n \geq N.$ Then if $\displaystyle \sum_{n=1}^{\infty}b_n$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_n$ also converges.
Suppose there exists an integer $N$ such that $a_n \geq b_n \geq 0$ for all $n \geq N.$ Then if $\displaystyle \sum_{n=1}^{\infty}b_n$ diverges, then $\displaystyle \sum_{n=1}^{\infty}a_n$ also diverges.
Back to Our Motivating Examples...
Thus, since $\displaystyle \frac{1}{n^2+1} \lt \frac{1}{n^2}$ for all $n \geq 1$ and $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ is known to converge, we may conclude that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2+1}$ also converges.
Also, since $\displaystyle \frac{1}{n} \lt \frac{1}{n-0.5}$ for all $n \geq 1$ and $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ is known to diverge, we may conclude that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n-0.5}$ also diverges.
As You Can See...
Using the Direct Comparison Test requires us to find a series which we know to converge or diverge.
Examples
$\displaystyle \sum_{n=1}^{\infty} \frac{5}{5n-1}$
The fact that $\displaystyle \frac{5}{5n-1}\approx \frac{5}{5n}=\frac{1}{n}$ for large $n$ suggests that the series diverges. Moreover,
it suggests that the harmonic series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ is a good choice for comparison.
Since $\displaystyle \frac{5}{5n-1}\geq \frac{5}{5n}=\frac{1}{n}$ for all $n \geq 1,$ and since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we conclude by the Direct Comparison Test that $\displaystyle \sum_{n=1}^{\infty} \frac{5}{5n-1}$ also diverges.
Since $\displaystyle \frac{5}{5n-1}\geq \frac{5}{5n}=\frac{1}{n}$ for all $n \geq 1,$ and since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we conclude by the Direct Comparison Test that $\displaystyle \sum_{n=1}^{\infty} \frac{5}{5n-1}$ also diverges.
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!}$
First, we need to find a series fo use for comparison.
Version 1: Compare to a $p$-Series. Here, we'll try the $p$-series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ which we know to be convergent.
We first note that $n!\geq n^2$ for all $n \geq 4.$ Then $\displaystyle 0\leq \frac{1}{n!}\leq \frac{1}{n^2}$ for all $n \geq 4.$
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ is known to converge, the above fact allows us to use the Direct Comparison Test to conclude that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!}$ also converges.
Scenic Alternative: Compare to a Geometric Series.
Since $$\displaystyle \frac{1}{n!}=1 \cdot \underbrace{\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{4}\cdots\frac{1}{n-1}\cdot\frac{1}{n}}_{\mbox{$n-1$ factors}}\leq \underbrace{\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdots\frac{1}{2}\cdot\frac{1}{2}}_{\mbox{$n-1$ copies}}=\frac{1}{2^{n-1}}=\left(\frac{1}{2}\right)^{n-1}$$ and since $\displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1}$ is known to converge, the Direct Comparison Test tells us that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!}$ also converges.
Version 1: Compare to a $p$-Series. Here, we'll try the $p$-series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ which we know to be convergent.
We first note that $n!\geq n^2$ for all $n \geq 4.$ Then $\displaystyle 0\leq \frac{1}{n!}\leq \frac{1}{n^2}$ for all $n \geq 4.$
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ is known to converge, the above fact allows us to use the Direct Comparison Test to conclude that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!}$ also converges.
Scenic Alternative: Compare to a Geometric Series.
Since $$\displaystyle \frac{1}{n!}=1 \cdot \underbrace{\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{4}\cdots\frac{1}{n-1}\cdot\frac{1}{n}}_{\mbox{$n-1$ factors}}\leq \underbrace{\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdots\frac{1}{2}\cdot\frac{1}{2}}_{\mbox{$n-1$ copies}}=\frac{1}{2^{n-1}}=\left(\frac{1}{2}\right)^{n-1}$$ and since $\displaystyle \sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1}$ is known to converge, the Direct Comparison Test tells us that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n!}$ also converges.
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n+\sqrt{n}}$
It is clear that $\displaystyle \frac{1}{2^n+\sqrt{n}} \lt \frac{1}{2^n}$ for all $n \geq 1.$
Therefore, since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n}$ converges, we have by the Direct Comparison Test that
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n+\sqrt{n}}$ also converges.
Another Motivating Example
Suppose we want to show that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n-1}$ converges.
The Direct Comparison Test doesn't apply.
What do we do?
The Limit Comparison Test
Suppose $a_n \geq 0$ and $b_n \geq 0$ for all $n \geq 1.$
i. If $\displaystyle \lim_{n \rightarrow \infty} \frac{a_n}{b_n}=L\neq 0$, then $\displaystyle \sum_{n=1}^{\infty} b_n$ and $\displaystyle \sum_{n=1}^{\infty} a_n$ both converge or both diverge.
ii. If $\displaystyle \lim_{n \rightarrow \infty} \frac{a_n}{b_n}=0$, and $\displaystyle \sum_{n=1}^{\infty} b_n$ converges, then $\displaystyle \sum_{n=1}^{\infty} a_n$ also converges.
iii. If $\displaystyle \lim_{n \rightarrow \infty} \frac{a_n}{b_n}=\infty$, and $\displaystyle \sum_{n=1}^{\infty} b_n$ diverges, then $\displaystyle \sum_{n=1}^{\infty} a_n$ also diverges.
Back to Our Motivating Example...
Use the Limit Comparison Test to show $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n-1}$ converges.
Use the geometric series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n}$ as the comparison.
Since $\displaystyle \frac{1}{2^n-1}\approx \frac{1}{2^n}$ for large $n,$ we will compare
the series $\displaystyle \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{1}{2^n-1}$ to a geometric series
$\displaystyle \sum_{n=1}^{\infty} b_n= \sum_{n=1}^{\infty}\frac{1}{2^n}$ by using the limit comparison test.
Now $$\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=\lim_{n \rightarrow \infty} \frac{\frac{1}{2^n-1}}{\frac{1}{2^n}}=\lim_{n \rightarrow \infty} \frac{2^n}{2^n-1}=1\neq 0.$$ Thus, by the Limit Comparison Test, since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n}$ converges, we know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n-1}$ also converges.
Now $$\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=\lim_{n \rightarrow \infty} \frac{\frac{1}{2^n-1}}{\frac{1}{2^n}}=\lim_{n \rightarrow \infty} \frac{2^n}{2^n-1}=1\neq 0.$$ Thus, by the Limit Comparison Test, since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n}$ converges, we know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^n-1}$ also converges.
More Examples!
$\displaystyle \sum_{n=1}^{\infty} \frac{2n+1}{(n+1)^2}$
Since $\displaystyle \frac{2n+1}{(n+1)^2}\approx \frac{2n}{n^2}=\frac{2}{n}$ for large $n,$ we will compare
the series $\displaystyle \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{2n+1}{(n+1)^2}$ to the harmonic series
$\displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty} \frac{1}{n}$ which is known to diverge.
Now $$\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=\lim_{n \rightarrow \infty} \frac{\frac{2n+1}{(n+1)^2}}{\frac{1}{n}}=\lim_{n \rightarrow \infty} \frac{2n^2+n}{(n+1)^2}=2\neq 0.$$ Thus, by the Limit Comparison Test, since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we know that $\displaystyle \sum_{n=1}^{\infty} \frac{2n+1}{(n+1)^2}$ also diverges.
Now $$\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=\lim_{n \rightarrow \infty} \frac{\frac{2n+1}{(n+1)^2}}{\frac{1}{n}}=\lim_{n \rightarrow \infty} \frac{2n^2+n}{(n+1)^2}=2\neq 0.$$ Thus, by the Limit Comparison Test, since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we know that $\displaystyle \sum_{n=1}^{\infty} \frac{2n+1}{(n+1)^2}$ also diverges.
$\displaystyle \sum_{n=1}^{\infty} \frac{1+n\ln n}{n^2+5}$
Since $\displaystyle \frac{1+n\ln n}{n^2+5}\approx \frac{n\ln n}{n^2}=\frac{\ln n}{n} \geq \frac{1}{n}$ for large $n,$ we will again compare
the series $\displaystyle \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{1+n\ln n}{n^2+5}$ to the harmonic series
$\displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty} \frac{1}{n}.$
Now $$\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=\lim_{n \rightarrow \infty} \frac{\frac{1+n\ln n}{n^2+5}}{\frac{1}{n}}=\lim_{n \rightarrow \infty} \frac{n+n^2\ln n}{n^2+5}=\infty$$ Thus, by the Limit Comparison Test, since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we know that $\displaystyle \sum_{n=1}^{\infty} \frac{1+n\ln n}{n^2+5}$ also diverges.
Now $$\lim_{n \rightarrow \infty} \frac{a_n}{b_n}=\lim_{n \rightarrow \infty} \frac{\frac{1+n\ln n}{n^2+5}}{\frac{1}{n}}=\lim_{n \rightarrow \infty} \frac{n+n^2\ln n}{n^2+5}=\infty$$ Thus, by the Limit Comparison Test, since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ diverges, we know that $\displaystyle \sum_{n=1}^{\infty} \frac{1+n\ln n}{n^2+5}$ also diverges.
$\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n^{3/2}}$
Finding a series here is a little trickier since the end behavior of $\displaystyle \frac{\ln n}{n^{3/2}}$ is not as clear cut as previous examples.
We first note that for sufficiently large $n,$ we have $\displaystyle \ln n \lt n^{1/4}.$ Thus, $\displaystyle \frac{\ln n}{n^{3/2}} \lt \frac{n^{1/4}}{n^{3/2}}=\frac{1}{n^{3/2-1/4}}=\frac{1}{n^{5/4}}$ for large $n.$ Thus, we will compare the series $\displaystyle \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{\ln n}{n^{3/2}}$ to the $p$-series $\displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty} \frac{1}{n^{5/4}}.$
Now $$ \begin{array}{lll} \displaystyle \lim_{n \rightarrow \infty} \frac{a_n}{b_n}&=\displaystyle \lim_{n \rightarrow \infty} \frac{\frac{\ln n}{n^{3/2}}}{\frac{1}{n^{5/4}}}&\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{n^{5/4}\ln n}{n^{3/2}}&\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{\ln n}{n^{3/2-5/4}}&\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{\ln n}{n^{1/4}}&\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{1/n}{\frac{1}{4}n^{-3/4}}&\mbox{L'Hopital!}\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{4n^{3/4}}{n}&\mbox{}\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{4}{n^{1/4}}&\mbox{}\\ &=0&\\ \end{array} $$ Thus, by the Limit Comparison Test, since the $p$-series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{5/4}}$ converges, we know that $\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n^{3/2}}$ also converges.
Special Note: The plain Comparison Test would have worked here too.
We first note that for sufficiently large $n,$ we have $\displaystyle \ln n \lt n^{1/4}.$ Thus, $\displaystyle \frac{\ln n}{n^{3/2}} \lt \frac{n^{1/4}}{n^{3/2}}=\frac{1}{n^{3/2-1/4}}=\frac{1}{n^{5/4}}$ for large $n.$ Thus, we will compare the series $\displaystyle \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{\ln n}{n^{3/2}}$ to the $p$-series $\displaystyle \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty} \frac{1}{n^{5/4}}.$
Now $$ \begin{array}{lll} \displaystyle \lim_{n \rightarrow \infty} \frac{a_n}{b_n}&=\displaystyle \lim_{n \rightarrow \infty} \frac{\frac{\ln n}{n^{3/2}}}{\frac{1}{n^{5/4}}}&\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{n^{5/4}\ln n}{n^{3/2}}&\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{\ln n}{n^{3/2-5/4}}&\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{\ln n}{n^{1/4}}&\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{1/n}{\frac{1}{4}n^{-3/4}}&\mbox{L'Hopital!}\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{4n^{3/4}}{n}&\mbox{}\\ &=\displaystyle \lim_{n \rightarrow \infty} \frac{4}{n^{1/4}}&\mbox{}\\ &=0&\\ \end{array} $$ Thus, by the Limit Comparison Test, since the $p$-series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{5/4}}$ converges, we know that $\displaystyle \sum_{n=1}^{\infty} \frac{\ln n}{n^{3/2}}$ also converges.
Special Note: The plain Comparison Test would have worked here too.