Integral Test definition formula proof and solved examples
Sequence and series are a building block of any analytical process. A function’s continuity can be easily proved with the help of sequences. In this respect, the integral test is used to find out whether a given series under analysis is in convergence or not. The convergence of series is important when the integral function has the sum of a series of a function. Hence, it is quite essential to check whether a series is in convergence or not in case of some specific functions.
If \[\int_{1}^{\infty} f(x) dx\] coverges, then \[\sum_{n=1}^{\infty} a^{n}\] converges.
If \[\int_{1}^{\infty} f(x) dx\] diverges, then \[\sum_{n=1}^{\infty} a^{n}\] diverges.
Integral Test for Convergence
In the mathematical domain, Integral test for convergence is a technique which is often applied for the purpose of testing an infinite series of non-negative terms for convergence. The method is also known as the Maclaurin-Cauchy test as Colin Maclaurin, and Augustin-Louis Cauchy developed it.
For example, if n is a neutral non-negative number, and it is a monotonically decreasing function, then the function is defined as
f: [n, ∞ ]→ R
In this case, the series will be convergent only if the integral is finite.
Integral Test Conditions
An integral comparison test is carried out mainly for integral terms. For instance, if there are two functions including f(x) and g(x) and g(x) ≥ f(x) on the given interval [c, ∞], then the following conditions should be true:
In case the term converges, then the term so does.
In case the term divergences, then the term so does.
Let’s understand the integral test series with the help of an example:
Example: Identify if the given series is convergent or divergent:
In this case, the function that will be used will be
F(x) = 1/ x In x
The function is positive, and if x is made larger, the denominator will be larger, and so the function will be decreasing. Thus, to determine the convergence of the following integral what needs to be done is
\[\int_{2}^{\infty} \frac{1}{x} ln x dx = \lim_{t \rightarrow \infty} \int_{2}^{t} \frac{1}{x} ln x dxu = lnx\]
\[= \lim_{t\rightarrow \infty} (ln(lnx))|_{2}^{t}\]
\[= \lim_{t\rightarrow \infty} (ln(lnt) - ln(ln2))\]
\[= \infty\]
This proves that the integral is divergent and so the series is divergent by the integral test.
Proof of the Integral Test
The integral test proof depends on the comparison test of the series. By now we know that is nothing but a sum of series \[\sum_{m}^{\infty} = N\int_{m+1}^{m} f(t)dt\].
Since “f” here is a monotonically decreasing function, then
f(t) ≤ f(m) for every “t” in [m, ∞]
For m > N, \[\int_{m+1}^{m} f(t)dt \leq \int_{m+1}^{m}f(m)dt = f(m)\]
It means that \[\int_{m+1}^{m} f(t)dt \leq f(m)\]
Since both the quantities are non-negative, a comparison test will be used.
If\[\sum_{m}^{\infty} = Nf(m)\] converges, then \[\sum_{m}^{\infty} = N\int_{m+1}^{m}f(t)dt = \int_{N}^{\infty} f(t)dt\] will also be convergent.
Thus, it is finite.
This, however, is just the first step of the proof.
Now, we again assume that the function f is a monotonically decreasing function, the equation that we get now is,
(m) ≤ f(t) for every “x” in [M, m]
So, \[f(m) = \int_{m-1}^{m} f(t)dt \leq \int_{m-1}^{m} f(t)dt\]
The comparison theorem proves that
If \[\sum_{m}^{\infty} = N\int_{m}^{m-1} f(t)dt = f(N) + \int_{N}^{\infty} f(t)dt\] converges, then we can say that \[\sum_{m}^{\infty} = Nf(m)\] will also converge. Hence, this proves the second part of the theorem.
The Comparison Test
If f(x) ≥ g(x) ≥ 0, on the interval [a,∞) then,
If \[\int_{a}^{\infty} f(x)dx\], converges then so does \[\int_{a}^{\infty}g(x)dx\].
\[\int_{a}^{\infty} g(x)dx\], diverges then so does \[\int_{a}^{\infty} f(x)dx\].
A comparison test makes a lot of sense when thinking in terms of area. If the function f(x) is larger than the function g(x) the area under the function f(x) will be larger than the area under the function g(x).
Thus, if the area under the larger function \[\int_{a}^{\infty} f(x)dx\] is finite and converges, then the area under the smaller function, i.e., \[\int_{a}^{\infty} g(x)dx\] will also converge and be finite.
Let’s take an example to understand this better.
Example: Find out whether
I = Converges or diverges
Solution: The function I does not have an expression in terms of elementary function. However
1 ≤ x ⇒ x ≤ x2
The last inequality follows as it is an increasing function
\[0 \leq = |_{1}^{\infty} = 1/e
Since 0 ≤ 1/e, the integral here converges.
FAQs on Integral Test for Series Convergence Explained
1. What is the Integral Test in calculus?
The Integral Test is a method used to determine whether an infinite series converges or diverges by comparing it to an improper integral. It applies to a series of the form ∑ aₙ where aₙ = f(n) and the function f(x) is:
- Positive
- Continuous
- Decreasing for x ≥ 1
If the improper integral ∫₁^∞ f(x) dx converges, then the series converges; if the integral diverges, the series diverges.
2. What are the conditions required to use the Integral Test?
The Integral Test requires the function corresponding to the series to be positive, continuous, and decreasing on [1, ∞). Specifically:
- aₙ = f(n)
- f(x) ≥ 0 for x ≥ 1
- f(x) is continuous
- f(x) is decreasing
If these conditions are satisfied, you can compare ∑ aₙ with ∫₁^∞ f(x) dx.
3. How do you use the Integral Test step by step?
To use the Integral Test, follow these steps:
- Identify aₙ and write it as a function f(x).
- Check that f(x) is positive, continuous, and decreasing.
- Evaluate the improper integral ∫₁^∞ f(x) dx.
- If the integral converges, the series converges; if it diverges, the series diverges.
This process determines convergence of infinite series using improper integrals.
4. Can you give an example of the Integral Test?
Yes, for the series ∑ 1/n², we apply the Integral Test using f(x) = 1/x².
- f(x) = 1/x² is positive, continuous, and decreasing for x ≥ 1.
- Compute ∫₁^∞ 1/x² dx.
- ∫ 1/x² dx = -1/x
- Lim (t→∞) [-1/x]₁ᵗ = 1
Since the integral converges to 1, the series ∑ 1/n² converges.
5. Does the Integral Test prove absolute convergence?
Yes, the Integral Test establishes convergence for series with positive terms, which implies absolute convergence. Because the test applies only when aₙ ≥ 0, if the integral converges, the series converges absolutely. It does not apply directly to alternating or sign-changing series.
6. What is the difference between the Integral Test and the Comparison Test?
The Integral Test compares a series to an improper integral, while the Comparison Test compares it to another known series.
- Integral Test: Uses ∫₁^∞ f(x) dx.
- Comparison Test: Compares aₙ with bₙ.
The Integral Test is especially useful for series involving powers like 1/nᵖ or logarithmic expressions.
7. When does the Integral Test fail?
The Integral Test fails when the function is not positive, not continuous, or not decreasing on [1, ∞). If any of these conditions are violated, the test cannot be applied. For example, oscillating or alternating series require other convergence tests such as the Alternating Series Test.
8. How is the Integral Test used for p-series?
The Integral Test shows that the p-series ∑ 1/nᵖ converges if p > 1 and diverges if p ≤ 1.
- Evaluate ∫₁^∞ 1/xᵖ dx.
- The integral equals 1/(p−1) when p > 1.
- The integral diverges when p ≤ 1.
This result is a key application of the Integral Test in calculus.
9. Why does the Integral Test work?
The Integral Test works because the area under a decreasing curve closely approximates the sum of rectangles representing the series terms. For a positive decreasing function:
- The series sum behaves like the improper integral.
- The integral provides upper and lower bounds for partial sums.
This geometric interpretation explains why convergence of ∫₁^∞ f(x) dx determines convergence of ∑ aₙ.
10. Can the Integral Test be used to estimate the remainder of a series?
Yes, the Integral Test provides a bound for the remainder (error) after n terms using the Integral Test remainder estimate.
- Rₙ ≤ ∫ₙ^∞ f(x) dx
- Rₙ ≥ ∫ₙ₊₁^∞ f(x) dx
This helps estimate how close a partial sum is to the total sum of a convergent infinite series.
