Integral Test for Convergence: A Comprehensive Guide with Examples
The concept of integral test on convergence may very well be one of the most valuable tests when it comes to calculus, of whether an infinite series is convergent or divergent. Integral test relates series to an improper integral and is systematic with regard to finding convergence of series, especially when dealing with nice functions.
This article presents the integral test definition to application, illustrate it using detailed examples, common pitfalls and the frequently asked question about the integral test, and how integral test compares with other convergence tests. Other important points which, however, are more obscure, such as visual intuition and relatedness to the real-world, will also be addressed.
What Is the Integral Test for Convergence?
Definition
Let f(n)=anf(n) = a_n be a positive, continuous, and decreasing function for n≥Nn \geq N. Then, the series ∑n=N∞an\sum_{n=N}^{\infty} a_n and the improper integral ∫N∞f(x) dx\int_N^{\infty} f(x)\,dx either both converge or both diverge.
Why It Works
A series can be pictured in the form of a sum of the heights of a rectangle. For example, the series ∑1n\sum \frac{1}{n} adds the heights of rectangles of width 1, stacked at each natural number. The area under the curve y=1xy = \frac{1}{x} from 1 to infinity approximates the total height. In case the area is infinite (divergent) then the sum has to be infinite.
One reason the integral test is so powerful is the relationship between definite integrals and infinite series; using tools and ideas of calculus to analyze sequences.
When Can You Use the Integral Test?
The function f(n)f(n), where an=f(n)a_n = f(n), must meet these conditions:
Positive: f(n)>0f(n) > 0 for all n≥Nn \geq N
Continuous: No discontinuities on [N,∞)[N, \infty)
Decreasing: f(n+1)≤f(n)f(n+1) \leq f(n) eventually
If any of these are not satisfied, you cannot apply the test directly.
Quick Checklist
Did it get a positive functional result?
Does it take place continuously?
Is it declining?
Are you able to estimate or evaluate the integral?If so, you are ok!
Step-by-Step Examples
1. Harmonic Series ∑1n\sum \frac{1}{n}
Let f(x)=1xf(x) = \frac{1}{x}. This is positive, continuous, decreasing for x≥1x \geq 1.
∫1∞1x dx=limt→∞lnt=∞\int_1^\infty \frac{1}{x} \, dx = \lim_{t \to \infty} \ln t = \infty
So the integral diverges ⇒ The series diverges. This shows that although the terms are diminishing to zero, the series does not converge.
2. P-Series ∑1np\sum \frac{1}{n^p}
Try p=2p = 2. Let f(x)=1x2f(x) = \frac{1}{x^2}.
∫1∞1x2 dx=[−1x]1∞=1\int_1^\infty \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^\infty = 1
The integral converges ⇒ The series converges. In general, ∑1np\sum \frac{1}{n^p} converges if and only if p>1p > 1.
3. Logarithmic Example: ∑1nlnn\sum \frac{1}{n \ln n}
Let f(x)=1xlnxf(x) = \frac{1}{x \ln x}, defined from x=2x = 2 to ∞\infty.
∫2∞1xlnx dx=ln(lnx)∣2∞=∞\int_2^\infty \frac{1}{x \ln x} \, dx = \ln(\ln x) \big|_2^\infty = \infty
The integral diverges ⇒ The series diverges.
4. Rational Example: ∑1n2+1\sum \frac{1}{n^2 + 1}
Let f(x)=1x2+1f(x) = \frac{1}{x^2 + 1}. This is smooth and decreasing for x≥1x \geq 1.
∫1∞1x2+1 dx=tan−1x∣1∞=π2−π4=π4\int_1^\infty \frac{1}{x^2 + 1} \, dx = \tan^{-1} x \big|_1^\infty = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}
Integral converges ⇒ Series converges.
Visual Intuition
Take f(x)=1xf(x) = 1/x. By plotting a curve and incrementing rectangles (width 1) on top of each of the natural numbers, we can see the following. The circumference in the curve between nn and n+1n+1 nearly approximates to the height of the rectangle in nn.
This graph assists in imagining the manner in which integrals act like sums across series. It is specifically useful academically.
Choosing Bounds & Handling Monotonicity
Can I Start at n=1n = 1?
Only if f(x)f(x) is decreasing from x=1x = 1. Otherwise, pick the point NN where it starts to decrease.
What If the Function Isn't Initially Decreasing?
As long as it's decreasing from some point onward, the test still applies. You just skip the first few terms, they don’t affect convergence.
When the Integral Test Fails or Isn’t Ideal
Not to Be Used:
The operation is discontinuous (e.g. vertical asymptote)
You can not easily combine it (e.g. inverse trig, exponential-log combinations)
It oscillates or goes down after a time.
When these occur, be a preference:
Comparison Test
Limit Comparison Test
Ratio/Root test of factorials/exponents
Real-World Example: Decaying Interest in Economics
Suppose a government gives a reward that reduces every year: $100,$50,$33.33,…$100, $50, $33.33, \ldots, resembling a harmonic series. Using the integral test, we see that the sum diverges, total payout is unbounded!
This can be modeled as ∑100n\sum \frac{100}{n}. Since the harmonic series diverges, so does the cumulative reward.
This shows how mathematical convergence impacts economic models.
Comparison with Other Convergence Tests
Common Pitfalls
Checking limit an→0a_n \to 0 is not enough to guarantee convergence
Applying test when function not decreasing
Using test on finite series
Ignoring domain where function becomes continuous or decreasing
Final Thoughts
When terms in an infinite series are similar to a known formula, such as 1/xp1/x^p, 1/(xlnx)1/(x \ln x), or 1/(x2+1)1/(x^2+1), the integral test provides us with a calculus perspective to study infinite series.
With the realization of when and how to apply it and its limitations, you will feel more confident to solve any problems concerning convergence.
FAQs
Q1. Can I use the integral test on ∑1lnn\sum \frac{1}{\ln n}? A: No, because ∫2∞1lnxdx\int_2^\infty \frac{1}{\ln x} dx diverges, and the terms don't decrease fast enough.
Q2. What if the function isn’t continuous at a single point? A: You can start the test from the next point where continuity begins.
Q3. Can I skip verifying monotonicity? A: Only if you’re certain or the function’s derivative confirms decreasing behavior.
Q4. Can I estimate the sum using the integral? A: Yes! Use bounds: ∫n∞f(x)dx<∑k=n∞f(k)<f(n)+∫n∞f(x)dx\int_n^{\infty} f(x)dx < \sum_{k=n}^\infty f(k) < f(n) + \int_n^{\infty} f(x)dx This is helpful for estimating error.
Q5. Does convergence mean the sum is small? A: Not always! A converging series can still sum to a large value, like ∑1n2=π26\sum \frac{1}{n^2} = \frac{\pi^2}{6}.
Pro Tip: Combine this test with comparison and visualization techniques for stronger problem-solving skills.
