The comparison test is an easy test for convergence or divergence when we know that, term by term, the terms of a test series are either greater than or less than those of our series of interest.
Here is a statement of the test in two parts, then we'll do some examples to illustrate it:
Here is the logic behind the first statement of the comparison test:
We get the convergent series used for comparison from our growing set of convergent series. For example, it might be a convergent geometric series or a convergent p-series.
The logic behind the second part of the comparison test goes like this:
We get the known divergent series used for comparison from among the many divergent series we've seen. For example, it might be a divergent geometric series (r < 1) or a divergent p-series (p < 1).
Does the series $\sum_{n = 1}^\infty \,\frac{1}{n!}$ converge?
$$\sum_{n = 1}^\infty \,\frac{1}{n^2}$$
the harmonic series, diverges, but that the p-series
$$\sum_{n = 1}^\infty \,\frac{1}{n}$$
converges. Now if each term of the series of interest is greater than $1/n$ for $n \gt 1$ then the series diverges, and if each term is less than $1/n^2$, then the series converges by the comparison test.
We know that for $n\ge 4$, $n!\gt n^2$, so each term of the series with $a_n=1/n!$ is less than the corresponding term of our convergent p-series, therefore the series converges. Here's a graphical look at these three series, $1/n$ in green, $1/n^2$ in
Now it's worth going back to the first green box above. The comparison test states that the terms of our test series ($1/n!$) must only be less, term-by-term, than a known convergent series for $n \gt N$, where $N$ is some integer. In this example, $N = 3$. For all $n\gt 3$, $1/n! \lt 1/n^2$, so the series converges. The extra stuff at the "front end" of the series (n = 1, 2, 3) is just a constant added on to a convergent series.
Does the series $\sum_{n = 1}^\infty \, \frac{\sqrt{n}}{n^2 + 5}$ converge?
$$\sum_{n = 1}^\infty \, \frac{\sqrt{n}}{n^2} = \sum_{n = 1}^\infty \frac{n^{1/2}}{n^2} = \sum_{n = 1}^\infty \,\frac{1}{n^{3/2}}$$
Now the series with terms $a_n = 1/n^{3/2}$ is a convergent p-series ($p = 3/2 \gt 1$). That suggests that we can show that each of the terms of our series is smaller than the corresponding term of the p-series with $p = 3/2$. First we set up the trial inequality:
$$\frac{\sqrt{n}}{n^2 + 5} \lt \frac{1}{n^{3/2}}$$
Cross multiplication gives:
$$n^{3/2} n^{1/2} \lt n^2 + 5$$
Using the laws of exponents on the left, we get:
$$n^2 \lt n^2 + 5$$
which reduces to
$$0 \lt 5$$
Because $0 \lt 5$ is always true, our series is, term-by-term, smaller than the convergent p-series with $a_n = 1/n^{3/2}$, so it converges.
Here (right column, top) is a table of the first ten terms of our series and the comparison series:
$n$ | $\frac{1}{n^{3/2}}$ | $\frac{\sqrt{n}}{n^2+5}$ |
---|---|---|
1 | 1.0000 | 0.1667 |
2 | 0.3536 | 0.1571 |
3 | 0.1925 | 0.1237 |
4 | 0.1250 | 0.0952 |
5 | 0.0894 | 0.0745 |
6 | 0.0680 | 0.0597 |
7 | 0.0540 | 0.0490 |
8 | 0.0442 | 0.0410 |
9 | 0.0370 | 0.0349 |
10 | 0.0316 | 0.0301 |
Below is a graphical representation of that comparison out to n = 30. While our the size of the terms of our function get closer to those of the comparison function, our algebra proves that they never cross.
Does the series $\sum_{n = 1}^\infty \,\frac{1}{n^2} cos \left(\frac{1}{n} \right)$ converge?
That means that this function is never greater than the convergent p-series with terms $a_n=1/n^2$, so the series converges by the comparison test.
Easy peasy, lemon-squeezy.
Does the series $\sum_{n = 1}^\infty \,\frac{n + 1}{\sqrt{n}}$ converge?
First let's break the sum into two parts.
$$\sum_{n = 1}^\infty \, \frac{n}{\sqrt{n}} + \frac{1}{\sqrt{n}} = \sum_{n = 1}^\infty \, n^{1/2} + \frac{1}{n^{1/2}}$$
Now the second is the divergent p-series with terms $a_n=1/n^{1/2}$ so the $n^{1/2}$ term just adds an ever-increasing amount to an already divergent series, therefore the series diverges.
There are a couple of things you should be aware of when using the comparison test.
Determine whether these series converge using the comparison test.
$$\sum_{n = 2}^\infty \, \frac{n^3}{n^4 - 1}$$
$$\sum_{n = 2}^\infty \, \frac{n^3}{n^4 - 1}$$
This series looks like the harmonic series, and the -1 in the denominator makes it look like it's greater, term-by-term, than the terms of that series. Let's compare them:
$$\frac{n^3}{n^4 - 1} \stackrel{?}{\gt} \frac{1}{n}$$
$$\longrightarrow \; n^4 \gt n^4 - 1$$
This is true for all n, so the series diverges; all of its terms are greater than the terms of the divergent harmonic series.
$$\sum_{n = 1}^\infty \, \frac{e^{1/n}}{n}$$
$$\sum_{n = 1}^\infty \, \frac{e^{1/n}}{n}$$
This series looks like the harmonic series, and the e1/n term in the numerator makes it look like it's greater, term-by-term, than that series. Let's compare
$$\frac{e^{1/n}}{n} \stackrel{?}{\gt} \frac{1}{n}$$
$$ \begin{align} \longrightarrow \; ne^{1/n} &\gt n \\[5pt] e^{1/n} &\gt 1 \\[5pt] e &\gt 1^n \end{align}$$
This is true for all $n$, so the series diverges; all of its terms are greater than the terms of the divergent harmonic series.
$$\sum_{n = 0}^\infty \, \frac{n + 5^n}{n + 6^n}$$
$$\sum_{n = 0}^\infty \, \frac{n + 5^n}{n + 6^n}$$
When n is large, the terms of the form "n" divide to 1. What's left is basically $\left( \frac{5}{6} \right)^n,$ which is a convergent geometric series. So let's try to show that, term-by-term, the terms of this series are less than those of that geometric series:
$$\frac{n + 5^n}{n+ 6^n} \stackrel{?}{\lt} \left( \frac{5}{6}\right)^n$$
$$ \begin{align} \longrightarrow \; n 6^n + 5^n6^n &\lt n 5^n + 6^{2n} \\[5pt] n(6^n - 5^n) &\lt 6n(6^n - 5^n) \\[5pt] n &\lt 6n \end{align}$$
This is true for all n > 0, so the series converges
$$\sum_{n = 0}^\infty \, \frac{2 + sin(n)}{10^n}$$
$$\sum_{n = 0}^\infty \, \frac{2 + sin(n)}{10^n}$$
The sin(n) term in the numerator is always within -1 < sin(n) < 1, so the numerator fluctuates between 1 and 3. If we view that numerator as a small constant, then this is nearly a geometric series with $r = \frac{1}{10}.$ Let's compare it to the convergent geometric series with $r = \frac{1}{2}.$
$$\frac{2 + sin(n)}{10^n} \stackrel{?}{\lt} \left( \frac{1}{2}\right)^n$$
$$ \begin{align} \longrightarrow \; 2 \cdot 2^n + 2^n sin(n) &\lt 10^n \\[5pt] 2^n (2 + sin(n)) &\lt 10^n \\[5pt] 2 + sin(n) &\lt 5^n \\[5pt] \end{align}$$
$$1 \lt 2 + sin(n) \lt 3$$
The inequality is true for all n ≥ 0, so the series converges.
$$\sum_{n = 1}^\infty \, \frac{4^{n + 1}}{3^n - 2}$$
$$\sum_{n = 1}^\infty \, \frac{4^{n + 1}}{3^n - 2}$$
For large values of n, the 2 has little effect, so this looks vaguely like the divergent harmonic series with $r = \frac{4}{3},$ and because the exponent in the numerator is one greater than that of the denominator, we further expect divergence. Compare this series to the divergent series with terms $\left(\frac{4}{3}\right)^n.$
$$\frac{4^{n + 1}}{3^n - 2} \stackrel{?}{\gt} \frac{4^n}{3^n}$$
$$ \begin{align} \longrightarrow \; 4^{n + 1}3^n &\gt 4^n(3^n - 2) \\[5pt] \frac{4^{n + 1}}{4^n} &\gt \frac{3^n - 2}{3^n} \\[5pt] 4 &\gt 1 - \frac{2}{3^n} \end{align}$$
This is true for all n ≥ 0, so the series diverges; all of its terms are greater than the terms of the divergent harmonic series.
$$\sum_{n = 1}^\infty \, \frac{n - 1}{n^2 \sqrt{n}}$$
$$\sum_{n = 1}^\infty \, \frac{n - 1}{n^2 \sqrt{n}}$$
If we disregard the 1 in the numerator, which has little effect for large n, the series has terms like (1/n)3/2, which is a convergent p-series, so compare this series to the p-seris with terms (1/n)3/2.
$$\frac{n - 1}{n^2 \sqrt{n}} \stackrel{?}{\lt} \left( \frac{1}{n} \right)^{3/2}$$
$$ \begin{align} \longrightarrow \; n\cdot n^{3/2} - n^{3/2} &\lt n^{5/2} \\[5pt] n - 1 &\lt n \end{align}$$
This is true for all n ≥ 1, so the series converges; all of its terms are less than the terms of the convergent harmonic series.
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