Series Testing Strategy
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 5.6: “Ratio and Root Tests” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/5-6-ratio-and-root-tests |
| Textbook used in class | Stewart, Calculus, Section 11.7: “Strategy for Testing Series” |
Opening Scenario
By the time you reach this section you have eight or nine tests available. The challenge is not remembering how each test works; it is recognizing which one to reach for. A decision procedure handles that: look at the form of $a_n$ and let the form point you to the test.
The Decision Procedure
Work through these questions in order. Stop at the first one that applies.
Step 1: Divergence Test (always check first). Compute $\lim_{n\to\infty} a_n$. If the limit is nonzero or does not exist, the series diverges immediately. Cost: almost nothing. If the limit is $0$, you know nothing yet -- move on.
Step 2: Recognize a standard form.
| If $a_n$ looks like... | Test to try |
|---|---|
| A geometric series $a r^n$ | Geometric series test: converges iff $|r| < 1$ |
| $1/n^p$ or a simple $p$-series | $p$-series: converges iff $p > 1$ |
| A rational function of $n$ (like $\frac{n^2+1}{n^3-2}$) | Limit Comparison with $1/n^{p}$, where $p$ = (denom degree) - (numer degree) |
| Contains $(-1)^n$ or $(-1)^{n+1}$ | Alternating Series Test |
| Contains $n!$ or a product with factorials | Ratio Test |
| Terms of the form $(b_n)^n$ | Root Test |
| Involves a continuous, positive, decreasing function | Integral Test |
Step 3: If the form is a rational function and the leading terms look like a $p$-series, use Limit Comparison. If the inequality is easy to write down, Direct Comparison is fine too.
Step 4: For mixed or unfamiliar forms, try Limit Comparison with the dominant-term approximation first. If that fails, try the Ratio Test.
Applied Examples
Example 1. $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n^2\sqrt{n}}$.
Step 1: $a_n \to 0$, so no conclusion yet. Step 2: Rational-looking. Dominant terms: $n/n^{5/2} = n^{-3/2}$, so try Limit Comparison with $1/n^{3/2}$.
$\lim \dfrac{(n-1)/(n^2\sqrt{n})}{1/n^{3/2}} = \lim \dfrac{(n-1)n^{3/2}}{n^{5/2}} = \lim \dfrac{n-1}{n} = 1$.
$\sum 1/n^{3/2}$ converges ($p = 3/2 > 1$). Converges.
Example 2. $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{5 \cdot 8 \cdot 11 \cdots (3n+2)}$.
Step 2: Contains $n!$: try Ratio Test.
$\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!}{5 \cdot 8 \cdots (3n+5)} \cdot \dfrac{5 \cdot 8 \cdots (3n+2)}{n!} = \dfrac{n+1}{3n+5} \to \dfrac{1}{3}$.
$L = 1/3 < 1$. Converges.
Example 3. $\displaystyle\sum_{n=1}^{\infty} \left(\frac{n^2+1}{2n^2+1}\right)^n$.
Step 2: Form $(b_n)^n$: try Root Test.
$|a_n|^{1/n} = \dfrac{n^2+1}{2n^2+1} \to \dfrac{1}{2} < 1$. Converges.
Example 4. $\displaystyle\sum_{n=2}^{\infty} \frac{1}{(\ln n)^2}$.
Step 1: Limit is $0$. Step 2: Not geometric, not a clean $p$-series, no factorial, no $(-1)^n$, no $n$-th power. Step 3: Since $(\ln n)^2 < n$ for large $n$, we have $1/(\ln n)^2 > 1/n$. Direct comparison with the divergent harmonic series: diverges.
trying the ratio test on every series. The ratio test gives $L = 1$ for any $p$-series, rational function, or polynomial ratio, making it useless for those cases. Students who reflexively apply the ratio test to everything waste time and end up stuck. The form of $a_n$ is the guide.
Common Patterns and Their Tests
| Pattern | Example | Test | Outcome |
|---|---|---|---|
| Geometric | $\sum (2/3)^n$ | Geometric | Converges, $r = 2/3$ |
| $p$-series | $\sum 1/n^3$ | $p$-series | Converges, $p = 3$ |
| Rational in $n$ | $\sum n/(n^2+1)$ | Limit Comparison with $1/n$ | Diverges |
| Alternating | $\sum (-1)^n/n$ | AST | Converges |
| Factorial | $\sum n^n/n!$ | Ratio | Diverges ($L = e > 1$) |
| $n$-th power | $\sum (n/(n+1))^n$ | Root | Converges ($L = 1/e < 1$) |
Leveled Practice
Problem 1. Which test would you use first for $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2+1}$? Apply it.
Show answer
The factor $(-1)^n$ signals the Alternating Series Test. $b_n = n/(n^2+1)$. Is $b_n$ decreasing? Let $f(x) = x/(x^2+1)$; $f'(x) = (1-x^2)/(x^2+1)^2 < 0$ for $x > 1$: yes, decreasing. And $b_n \to 0$. Converges by AST.
Problem 2. Which test for $\displaystyle\sum_{n=1}^{\infty} \frac{2^n n!}{(2n)!}$?
Show answer
Factorial: Ratio Test. $\dfrac{a_{n+1}}{a_n} = \dfrac{2^{n+1}(n+1)!}{(2n+2)!} \cdot \dfrac{(2n)!}{2^n n!} = \dfrac{2(n+1)}{(2n+2)(2n+1)} = \dfrac{1}{2n+1} \to 0 < 1$. Converges.