Comparison Tests for Series

Is $\sum \frac{1}{n^2 + 1}$ convergent? It looks a lot like $\sum \frac{1}{n^2}$, which you know converges. The comparison tests turn that “looks like” into a proper argument. You compare a new series with a p-series or a geometric series whose behaviour you already know.

Key ideas

Both tests on this page are for series with positive terms (at least eventually).

The direct comparison test

Suppose $0 \le a_n \le b_n$ for all $n$ (or for all $n$ from some point on).

• If $\sum b_n$ converges, then $\sum a_n$ converges. (Smaller than something finite is finite.)
• If $\sum a_n$ diverges, then $\sum b_n$ diverges. (Bigger than something infinite is infinite.)

The other two directions tell you nothing. Being smaller than a divergent series, or bigger than a convergent one, proves nothing.

Every term $\frac{1}{n^2 + 1}$ is smaller than $\frac{1}{n^2}$. Since $\sum \frac{1}{n^2}$ converges, so does $\sum \frac{1}{n^2 + 1}$.

The limit comparison test

Sometimes the inequality goes the wrong way, or is messy to prove. Then compare the terms with a limit instead. If $a_n \gt 0$ and $b_n \gt 0$ and

$$\lim_{n \to \infty} \frac{a_n}{b_n} = L, \quad\text{where } 0 \lt L \lt \infty \text{ (a positive, finite number)},$$

then $\sum a_n$ and $\sum b_n$ both converge or both diverge. The idea: for large $n$, $a_n \approx L\,b_n$, and multiplying by a constant doesn’t change convergence.

If the limit is $0$ or $\infty$, this form of the test doesn’t decide anything. Pick a different $b_n$.

Choosing what to compare with

Keep only the dominant parts of the top and bottom: the highest power of $n$ in a polynomial, or the biggest exponential. For example,

$$\frac{3n + 1}{n^3 - 2n + 4} \;\text{ behaves like }\; \frac{3n}{n^3} = \frac{3}{n^2}, \quad\text{so compare with } b_n = \frac{1}{n^2} .$$

Useful facts for direct comparison: $0 \le \sin^2 n \le 1$, $0 \le \cos^2 n \le 1$, and $\ln n \lt n$.

On the AP exam, show the comparison explicitly: state the inequality (for direct comparison) or the limit and its value (for limit comparison), name the known series, and say what it does.

Worked examples

Example 1: Direct comparison, convergent

Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converge or diverge?

Solution. For every $n \ge 1$, $n^2 + 1 \gt n^2$, so

$$0 \lt \frac{1}{n^2 + 1} \lt \frac{1}{n^2} .$$

$\sum \frac{1}{n^2}$ is a convergent p-series ($p = 2 \gt 1$). Our series is smaller term by term, so it converges by the direct comparison test.

Example 2: Direct comparison, divergent

Does $\displaystyle\sum_{n=2}^{\infty} \frac{1}{\ln n}$ converge or diverge?

Solution. For $n \ge 2$, $0 \lt \ln n \lt n$, so

$$\frac{1}{\ln n} \gt \frac{1}{n} .$$

$\sum \frac{1}{n}$ is the harmonic series, which diverges. Our series is bigger term by term, so it diverges by the direct comparison test.

Example 3: Limit comparison

Does $\displaystyle\sum_{n=1}^{\infty} \frac{3n + 1}{n^3 - 2n + 4}$ converge or diverge?

Solution. The denominator is positive for $n \ge 1$ (its values start $3, 8, 25, \dots$ and keep growing), so the terms are positive. The dominant parts give $\frac{3n}{n^3} = \frac{3}{n^2}$, so compare with $b_n = \dfrac{1}{n^2}$:

$$\lim_{n \to \infty} \frac{\dfrac{3n + 1}{n^3 - 2n + 4}}{\dfrac{1}{n^2}} = \lim_{n \to \infty} \frac{3n^3 + n^2}{n^3 - 2n + 4} = 3$$

Since $0 \lt 3 \lt \infty$ and $\sum \frac{1}{n^2}$ converges ($p = 2$), the series converges by the limit comparison test.

(A direct comparison would be awkward here: $\frac{3n + 1}{n^3 - 2n + 4}$ is bigger than $\frac{1}{n^2}$, which is the useless direction.)

Example 4: Comparing with a geometric series

Does each series converge or diverge?

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{3^n - 1}$

Solution.

(a) Since $0 \le \sin^2 n \le 1$, we have $0 \le \dfrac{\sin^2 n}{n^2} \le \dfrac{1}{n^2}$. The p-series $\sum \frac{1}{n^2}$ converges, so this series converges by direct comparison. ($\sin n$ is in radians here, but the bound works either way.)

(b) The dominant parts give $\frac{2^n}{3^n} = \left(\frac{2}{3}\right)^n$, a convergent geometric series. The terms are a bit bigger than $\left(\frac{2}{3}\right)^n$, so use the limit comparison test:

$$\lim_{n \to \infty} \frac{\dfrac{2^n}{3^n - 1}}{\dfrac{2^n}{3^n}} = \lim_{n \to \infty} \frac{3^n}{3^n - 1} = \lim_{n \to \infty} \frac{1}{1 - 3^{-n}} = 1$$

Since $0 \lt 1 \lt \infty$ and $\sum \left(\frac{2}{3}\right)^n$ converges (geometric, $\lvert r \rvert \lt 1$), the series converges by the limit comparison test.

Common mistakes

Comparing in the useless direction. ”$\frac{1}{n^2 - 1} \gt \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges” proves nothing. For convergence you need to be smaller than a convergent series; for divergence, bigger than a divergent one. If the inequality goes the wrong way, switch to the limit comparison test.

Forgetting the terms must be positive. Both tests need $a_n \gt 0$ (at least eventually). For a series like $\sum \frac{\sin n}{n^2}$ with mixed signs, compare $\lvert a_n \rvert$ instead (see absolute convergence).

Getting L = 0 or infinity and drawing a conclusion anyway. In the AP form of the limit comparison test, $L$ must be a positive finite number. If you get $0$ or $\infty$, you probably chose the wrong $b_n$.

Comparing with the wrong series. For $\frac{n + 1}{n^3}$, the dominant behaviour is $\frac{n}{n^3} = \frac{1}{n^2}$, not $\frac{1}{n^3}$. Cancel the powers carefully before choosing $b_n$.

Leaving out the justification. “It’s like $\frac{1}{n^2}$, so it converges” isn’t enough. Write the inequality or the limit, name the comparison series, and state the test.

Practice

1. (Warm-up) Use direct comparison to decide whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^3 + 7}$ converges.

Solution

$n^3 + 7 \gt n^3$, so $0 \lt \dfrac{1}{n^3 + 7} \lt \dfrac{1}{n^3}$. The p-series $\sum \frac{1}{n^3}$ converges ($p = 3$), so the series converges by direct comparison.

2. (Warm-up) Use direct comparison to decide whether $\displaystyle\sum_{n=1}^{\infty} \frac{n + 1}{n^2}$ converges.

Solution

$\dfrac{n + 1}{n^2} = \dfrac{1}{n} + \dfrac{1}{n^2} \gt \dfrac{1}{n}$. The harmonic series diverges, and our series is bigger, so it diverges by direct comparison.

3. (Warm-up) Use direct comparison to decide whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{3^n + 2}$ converges.

Solution

$3^n + 2 \gt 3^n$, so $0 \lt \dfrac{1}{3^n + 2} \lt \left(\dfrac{1}{3}\right)^n$. The geometric series $\sum \left(\frac{1}{3}\right)^n$ converges ($\lvert r \rvert \lt 1$), so the series converges by direct comparison.

4. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{n^2 + 1}{n^4 + n}$ converge or diverge?

Solution

The dominant parts give $\frac{n^2}{n^4} = \frac{1}{n^2}$. Limit comparison with $b_n = \frac{1}{n^2}$:

$$\lim_{n \to \infty} \frac{n^2 + 1}{n^4 + n} \cdot n^2 = \lim_{n \to \infty} \frac{n^4 + n^2}{n^4 + n} = 1$$

Since $0 \lt 1 \lt \infty$ and $\sum \frac{1}{n^2}$ converges, the series converges by the limit comparison test.

5. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 4}}$ converge or diverge?

Solution

For large $n$, $\sqrt{n^2 + 4} \approx n$, so compare with $b_n = \frac{1}{n}$:

$$\lim_{n \to \infty} \frac{\dfrac{1}{\sqrt{n^2 + 4}}}{\dfrac{1}{n}} = \lim_{n \to \infty} \frac{n}{\sqrt{n^2 + 4}} = \lim_{n \to \infty} \frac{1}{\sqrt{1 + 4/n^2}} = 1$$

Since $0 \lt 1 \lt \infty$ and the harmonic series diverges, the series diverges by the limit comparison test.

6. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{\cos^2 n}{n\sqrt{n}}$ converge or diverge?

Solution

Since $0 \le \cos^2 n \le 1$ and $n\sqrt{n} = n^{3/2}$:

$$0 \le \frac{\cos^2 n}{n\sqrt{n}} \le \frac{1}{n^{3/2}}$$

The p-series $\sum \frac{1}{n^{3/2}}$ converges ($p = \tfrac{3}{2} \gt 1$), so the series converges by direct comparison.

7. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{\ln n}{n^3}$ converge or diverge?

Solution

For $n \ge 1$, $0 \le \ln n \lt n$, so

$$0 \le \frac{\ln n}{n^3} \lt \frac{n}{n^3} = \frac{1}{n^2}$$

The p-series $\sum \frac{1}{n^2}$ converges, so the series converges by direct comparison. (The first term is $0$; that’s fine.)

8. (Challenge) Does $\displaystyle\sum_{n=1}^{\infty} \sin\!\left(\frac{1}{n}\right)$ converge or diverge?

Solution

For $n \ge 1$, $0 \lt \frac{1}{n} \le 1 \lt \pi$, so the terms are positive. Compare with $b_n = \frac{1}{n}$. Let $t = \frac{1}{n}$, which goes to $0^+$:

$$\lim_{n \to \infty} \frac{\sin(1/n)}{1/n} = \lim_{t \to 0^+} \frac{\sin t}{t} = 1$$

Since $0 \lt 1 \lt \infty$ and the harmonic series diverges, the series diverges by the limit comparison test.

9. (Challenge) Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1 + 1/n}}$ converge or diverge? (Careful: the exponent is more than $1$, but it changes with $n$.)

Solution

This is not a p-series, because the exponent isn’t constant. Write $\dfrac{1}{n^{1 + 1/n}} = \dfrac{1}{n \cdot n^{1/n}}$ and compare with $b_n = \frac{1}{n}$:

$$\lim_{n \to \infty} \frac{\dfrac{1}{n \cdot n^{1/n}}}{\dfrac{1}{n}} = \lim_{n \to \infty} \frac{1}{n^{1/n}}$$

To find $\lim n^{1/n}$, take logs: $\ln\!\left(n^{1/n}\right) = \dfrac{\ln n}{n} \to 0$, so $n^{1/n} \to e^0 = 1$. The limit is $1$.

Since $0 \lt 1 \lt \infty$ and the harmonic series diverges, the series diverges by the limit comparison test. The exponent creeps down toward $1$ too quickly to save it.