Series Convergence Tests Practice Problems with Answers
12 series convergence practice problems with answers: nth term, p-series, geometric, comparison, integral, ratio, root and alternating series tests.
Sequences and series Updated 3 min read
At a glance
- Problems
- 12, with answers and full solutions
- Level
- Calculus 2 and AP Calculus BC, Unit 10
- You need
- Limits at infinity and the geometric series formula
- Main idea
- Match the shape of the terms to the test that handles that shape
- Practice with new numbers
- Choosing a test lesson
On this page
These series convergence test practice problems give you a mix of series and ask you to decide whether each one converges or diverges, and why. Picking the right test is the real skill, so try to name the test before you read the solution. The answers come first, then the full solutions.
A quick guide to the tests:
| Shape of the terms | Test to try |
|---|---|
| Terms do not go to 0 | nth term test: diverges |
| ar^n | Geometric: converges if \lvert r\rvert < 1 |
| \frac{1}{n^p} | p-series: converges if p > 1 |
| Rational or root expressions | Comparison or limit comparison |
| Factorials or c^n | Ratio test |
| (\dots)^n | Root test |
| (-1)^n b_n | Alternating series test |
Problems
Decide whether each series converges or diverges.
Benchmarks
- \displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3}
- \displaystyle\sum_{n=1}^{\infty}\frac{n}{n + 1}
- \displaystyle\sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^n (find the sum too)
- \displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}
Comparison and integral tests
- \displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2 + 1}
- \displaystyle\sum_{n=1}^{\infty}\frac{1}{2n - 1}
- \displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}
Ratio and root tests
- \displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}
- \displaystyle\sum_{n=1}^{\infty}\frac{n!}{10^n}
- \displaystyle\sum_{n=1}^{\infty}\left(\frac{n}{2n + 1}\right)^n
Alternating series
- \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}
- \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^2}
Answers
| Problem | Answer | Test |
|---|---|---|
| 1 | Converges | p-series, p = 3 |
| 2 | Diverges | nth term, terms \to 1 |
| 3 | Converges to 3 | Geometric, r = \frac{2}{3} |
| 4 | Diverges | p-series, p = \frac{1}{2} |
| 5 | Converges | Comparison with \frac{1}{n^2} |
| 6 | Diverges | Limit comparison with \frac{1}{n} |
| 7 | Diverges | Integral test |
| 8 | Converges | Ratio test, L = \frac{1}{2} |
| 9 | Diverges | Ratio test, L = \infty |
| 10 | Converges | Root test, L = \frac{1}{2} |
| 11 | Converges conditionally | Alternating series test |
| 12 | Converges absolutely | \sum\frac{1}{n^2} converges |
Step-by-step solutions
1. This is a p-series with p = 3 > 1, so it converges.
2. The terms approach 1, not 0:
By the nth term test, the series diverges.
3. Geometric with first term 1 and ratio \frac{2}{3}. Since |r| < 1:
4. \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}} is a p-series with p = \frac{1}{2} \le 1, so it diverges.
5. For all n, n^2 + 1 > n^2, so \frac{1}{n^2 + 1} < \frac{1}{n^2}. The bigger series \sum\frac{1}{n^2} converges, so the smaller one does too.
6. Compare with \frac{1}{n}:
The limit is positive and finite, and \sum\frac{1}{n} diverges, so this series diverges too.
7. The function \frac{1}{x\ln x} is positive and decreasing for x \ge 2. With u = \ln x:
The integral diverges, so the series diverges.
8. Ratio test:
It converges. In fact, its sum is 2.
9. Ratio test:
It diverges. Factorials grow faster than any exponential.
10. Root test:
It converges.
11. The terms \frac{1}{n} are positive, decreasing and go to 0, so the alternating series test says it converges. But \sum\frac{1}{n} diverges, so the convergence is conditional. The sum is \ln 2.
12. The absolute values form \sum\frac{1}{n^2}, a convergent p-series. So the series converges absolutely.
Common mistakes on these problems
- Using the nth term test to prove convergence. Terms going to 0 does not show convergence. Problem 7 has terms going to 0 and still diverges.
- Comparing in the wrong direction. Being smaller than a divergent series, or bigger than a convergent one, proves nothing.
- Using the ratio test on p-series. For \frac{1}{n^p} the ratio limit is always 1, which gives no answer.
- Forgetting to check absolute convergence. For alternating series, the full answer says whether the convergence is absolute or conditional.
Keep practicing
For unlimited series with instant feedback, try the choosing a test lesson, the ratio and root tests lesson and the series practice set. For the reasoning behind each test, read the convergence tests guide. Then use these tests to find intervals in the power series problems.
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Choosing a test
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