Comparison Test Calculator
Enter the general term of a series and a comparison series to test for convergence or divergence using the Limit Comparison Test.
How to use this calculator
Type the general term of your series, using n as the variable. For example, enter 1/(n^2+3). You can write 2n instead of 2*n, and you can use sqrt(), ln(), exp(), sin(), cos(), abs(), pi and e.
Next, pick a series to compare against. Choose a p-series (1/n^p) or a geometric series (r^n) and enter p or r. Press Calculate. You will see the estimated limit of the ratio and a plain-language conclusion. Press Reset to start over.
The terms of your series must be positive. If a(n) is negative for large n, the calculator tells you so instead of returning a wrong answer.
What this calculator does
It applies the limit comparison test. It divides your series term a(n) by the comparison term b(n). Then it estimates what that ratio approaches as n grows very large.
The estimate is numerical. The tool evaluates the ratio at very large values of n and checks whether it settles on a number, shrinks to zero, or grows without bound. This is an estimate, not a proof. When the ratio moves too slowly to read, the tool says it could not be determined instead of guessing.
What is the comparison test for series?
It decides whether an infinite series converges or diverges by comparing it with a series you already understand. There are two versions.
The direct comparison test compares the terms one by one. If 0 ≤ a(n) ≤ b(n) and the series of b(n) converges, then the series of a(n) converges. If a(n) ≥ b(n) ≥ 0 and the series of b(n) diverges, then the series of a(n) diverges.
The limit comparison test looks at the limit of a(n) / b(n) instead. It is easier to apply, because you do not have to prove an inequality for every term. That is why this calculator uses it.
Series with known behavior
Pick a comparison series whose behavior you already know. These are the most useful ones.
| Series | Form | Converges when |
|---|---|---|
| p-series | 1 / n^p | p > 1 |
| Geometric series | r^n | |r| < 1 |
| Harmonic series | 1 / n (p = 1) | Never. It diverges. |
How the limit comparison test works
Find L, the limit of a(n) / b(n) as n approaches infinity. Then use this table.
| Limit L | If b(n) converges | If b(n) diverges |
|---|---|---|
| 0 < L < ∞ | a(n) converges | a(n) diverges |
| L = 0 | a(n) converges | Inconclusive |
| L = ∞ | Inconclusive | a(n) diverges |
When L is a positive, finite number, the two series behave alike. That is the most common and most useful case.
How to choose the comparison series
Keep only the dominant terms of a(n) and ignore the rest. For a fraction of polynomials, compare the highest powers.
So compare with the p-series 1 / n², which has p = 2.
Rule of thumb: if the denominator has degree D and the numerator has degree d, compare with 1 / n^(D − d). The series converges when D − d is greater than 1.
Worked examples
The ratio n² / (n² + 3) approaches 1. The limit is finite and positive, and the p-series with p = 2 converges. So the series converges.
The ratio approaches 2. Again the limit is finite and positive, and p = 2 converges. So the series converges.
The ratio √n / (√n + 1) approaches 1. The p-series with p = 0.5 diverges, because p is not greater than 1. So this series diverges too.
The ratio (n + 1) / 1.5ⁿ approaches 0, because the exponential grows faster than n. The geometric series with r = 0.5 converges. With L = 0 and a convergent comparison series, the series converges.
What to do when the result is inconclusive
An inconclusive result does not mean the series has no answer. It means this comparison did not settle it. Try a different comparison series.
The ratio is n, which approaches ∞. The comparison series converges, so the test is inconclusive. Now compare with p = 1 instead. The ratio approaches 1, and the p-series with p = 1 diverges. So 1 / n diverges.
Other convergence tests
Some series need a different tool. Here is when to reach for each one.
Integral test. Use it when the terms are positive and decreasing, and you can integrate the matching function. The series 1 / (n ln n) is a good case. Its ratio to a p-series moves too slowly for a numeric estimate, but the integral test shows that it diverges.
Ratio test. Use it for factorials and exponentials. Find the limit of a(n+1) / a(n). A limit below 1 means convergence, and a limit above 1 means divergence.
Root test. Use it when the term is raised to the power n.
Alternating series test. Use it when the signs alternate and the terms shrink to zero.
Divergence test (nth-term test). If the terms do not approach zero, the series diverges. Check this first.
Common mistakes
Treating L = 0 as always convergent. It only works when the comparison series converges.
Treating p = 1 as convergent. The harmonic series diverges. A p-series needs p greater than 1.
Using negative terms. The test needs positive terms. Series with mixed signs need other tests.
Comparing the wrong way in the direct test. Being smaller than a divergent series proves nothing. Being smaller than a convergent series proves convergence.
Frequently asked questions
Why does a p-series need p greater than 1?
The terms of 1 / n^p must shrink fast enough for the total to settle. When p is greater than 1, they do. When p is 1 or less, the sum grows without bound.
Why does a geometric series need |r| less than 1?
Each term is r times the one before it. If |r| is less than 1, the terms shrink toward zero quickly enough for the sum to settle. If |r| is 1 or more, the terms stay the same size or grow.
What does inconclusive mean?
The limit did not fit a case that the test can decide. Choose a different comparison series, or use another test such as the integral test or the ratio test.
Does the starting index of the series matter?
Not for convergence. Adding or removing a finite number of terms changes the sum, but it does not change whether the series converges.
How accurate is the limit estimate?
It works well for rational expressions, roots, and exponentials. Limits that move very slowly, such as ones involving ln(n), may come back as not determined. In that case, work the limit by hand or try another test.
Can I use series with negative terms?
No. The test needs positive terms. For series with mixed signs, test the absolute values first or use the alternating series test.