Integral Test Calculator: Determine Series Convergence with Step-by-Step Solutions
The integral test calculator determines whether an infinite series converges or diverges by evaluating the corresponding improper integral ∫n₀∞ f(x) dx. Enter your function, and the calculator verifies all three conditions (positive, decreasing, continuous), computes the integral, and delivers a step-by-step convergence verdict — including remainder estimates so you know how accurate your partial sums are.
Whether you need to check a p-series like Σ1/n², a logarithmic series like Σ1/(n·ln n), or any custom function, this integral convergence calculator handles both analytic closed-form solutions and numerical approximations for general functions. Every result includes detailed integration steps showing substitutions, limit evaluations, and the final convergence conclusion.
How to Use the Integral Test Calculator with Steps
Using the integral test calculator with steps takes just three inputs: the function f(x), the starting index n₀, and an optional remainder index N for error bounds. Here is exactly how the tool works:
- Enter f(x) — type the function whose terms define your series, such as
1/x^2,1/(x*ln(x)), ore^(-x). - Set n₀ — choose the starting index (usually 1 or 2 for logarithmic functions).
- Click "Apply Test" — the calculator checks that f(x) is positive, decreasing, and continuous, then evaluates the improper integral.
- Read the verdict — see whether the series converges or diverges, with the exact integral value and step-by-step working.
The calculator first tries to recognize your function analytically (p-series, logarithmic, exponential, arctangent patterns). If the function doesn't match a known pattern, it falls back to numerical integration using Simpson's rule with progressive upper bounds to determine convergence — so virtually any integrable function can be tested.
What Is the Integral Test for Series Convergence?
The integral test (also called the Maclaurin–Cauchy test) is a method for determining whether an infinite series Σn=n₀∞ f(n) converges or diverges. The test compares the discrete sum to a continuous improper integral.
Integral Test Theorem
Let f(x) be a function that is positive, continuous, and decreasing for all x ≥ n₀. Then the series Σn=n₀∞ f(n) and the improper integral ∫n₀∞ f(x) dx either both converge or both diverge.
Three Conditions for the Integral Test
Before applying the integral test, you must verify that f(x) satisfies all three conditions on the interval [n₀, ∞):
- Positive: f(x) > 0 for all x ≥ n₀. The function values (and therefore the series terms) must be strictly positive.
- Continuous: f(x) has no breaks, jumps, or vertical asymptotes on [n₀, ∞).
- Decreasing: f(x) is a monotonically decreasing function, meaning f'(x) < 0 for x ≥ n₀. The terms must get smaller as n increases.
Our calculator numerically samples f(x) at multiple points to verify the positive and decreasing conditions automatically, flagging any violations before reporting the result.
Integral Test Examples with Solutions
Example 1: p-Series Σ1/n² (Converges)
Consider the series Σn=1∞ 1/n². Set f(x) = 1/x². This function is positive, continuous, and decreasing for x ≥ 1.
∫1∞ 1/x² dx = [−1/x]1∞ = 0 − (−1) = 1
Since the integral equals 1 (finite), the series converges.
Example 2: Harmonic Series Σ1/n (Diverges)
The harmonic series Σn=1∞ 1/n corresponds to f(x) = 1/x. All three conditions are satisfied for x ≥ 1.
∫1∞ 1/x dx = [ln(x)]1∞ = ∞ − 0 = ∞
Since the integral is infinite, the harmonic series diverges.
Example 3: Logarithmic Series Σ1/(n·ln n) (Diverges)
For the series Σn=2∞ 1/(n·ln n), set f(x) = 1/(x·ln x). Note we start at n = 2 since ln(1) = 0.
Substitution: u = ln(x), du = dx/x
∫2∞ 1/(x·ln x) dx = ∫ln 2∞ 1/u du = [ln(u)]ln 2∞ = ∞
The integral diverges, so Σ1/(n·ln n) diverges.
Integral Test vs. Other Convergence Tests
The integral test is one of several convergence tests for infinite series. Here is when to use the integral test compared to other methods:
| Test | Best For | Limitation |
|---|---|---|
| Integral Test | p-series, logarithmic series, functions easy to integrate | Requires positive, decreasing, continuous f(x) |
| Ratio Test | Series with factorials, exponentials, n! | Inconclusive when L = 1 |
| Root Test | Series with n-th powers like (aₙ)^n | Inconclusive when L = 1 |
| Comparison Test | When a simpler bounding series is obvious | Need to know a comparison series |
| Alternating Series | Series with alternating signs (−1)^n | Only works for alternating series |
The integral test is often the first choice for series where the general term f(n) corresponds to a function that is straightforward to integrate. For series involving factorials or ratios of successive terms, the ratio test calculator is usually more efficient. Use our series convergence calculator to automatically determine which test to apply.
Integral Test Remainder Estimate
When the integral test confirms convergence, it also provides remainder estimates (error bounds) for partial sums. If S = Σn=1∞ f(n) and SN = Σn=1N f(n), then the remainder RN = S − SN satisfies:
∫N+1∞ f(x) dx ≤ RN ≤ ∫N∞ f(x) dx
This tells you exactly how close your partial sum SN is to the true infinite series value. Our calculator computes the upper bound ∫N∞ f(x) dx for any N you specify, so you can determine how many terms are needed for a given accuracy.
Common Series and Their Convergence by the Integral Test
The p-Series Test
The p-series Σn=1∞ 1/np is the most important application of the integral test. Using f(x) = 1/xp:
- p > 1: ∫1∞ 1/xp dx = 1/(p−1), which is finite → series converges
- p = 1: ∫1∞ 1/x dx = ln(∞) = ∞ → harmonic series diverges
- p < 1: ∫1∞ 1/xp dx = ∞ → series diverges
Try it yourself: enter 1/x^3 in the calculator to see the convergence proof for Σ1/n³.
Logarithmic Series
Series involving natural logarithms are a classic application of the integral test because the substitution u = ln(x) transforms the integral into a simpler p-integral:
- Σ1/(n·ln n): Diverges (equivalent to ∫ 1/u du after substitution)
- Σ1/(n·(ln n)²): Converges (equivalent to ∫ 1/u² du)
- Σ1/(n·(ln n)p): Converges if and only if p > 1
When Should You Use the Integral Test?
The integral test is the right choice when:
- ✓The series terms aₙ = f(n) come from a function f(x) that you can integrate (polynomials, logarithms, exponentials)
- ✓You are dealing with p-series (Σ1/np) or logarithmic series (Σ1/(n·(ln n)p))
- ✓You need a remainder estimate to know how many terms give a desired accuracy
- ✓The ratio test and root test are inconclusive (L = 1), which often happens with polynomial-type terms
The integral test does not work when f(x) is not positive, not decreasing, or not continuous. For alternating series like Σ(−1)n/n, use the alternating series test calculator instead.
About the Author
Why Students and Educators Trust This Integral Test Calculator
Our integral test calculator combines the speed of pattern recognition with the versatility of numerical methods. For standard series types (p-series, logarithmic, exponential), you get instant closed-form results with exact integration steps. For non-standard functions, the numerical engine uses Simpson's rule with adaptive bounds to approximate the improper integral and assess convergence.
Every calculation includes condition verification, step-by-step integration, the convergence/divergence verdict, and remainder bounds — everything you need for homework, exam prep, or professional numerical analysis. The calculator works seamlessly on both desktop and mobile devices, making it your reliable companion whenever you encounter series convergence problems.
For a complete convergence testing workflow, pair this tool with our series convergence calculator (which runs multiple tests automatically), the ratio test calculator for factorial/exponential series, and the definite integral calculator for detailed integration practice.
