# Integral Test

Sequences and series act as a building block for the analysis process. The continuity of the function can be easily proved using the sequences. The integral test is used to find whether the given series is convergence or not. The convergence of series is more significant in many situations when the integral function has the sum of a series of functions. So, it is essential to check whether the given series is convergence or not while dealing with some specific functions on sequences and series. In this article, let us have a look at the procedure of integral test, proof and comparison test.

## Integral Test for Convergence

The integral test for convergence is a method used to test the infinite series of non-negative terms for convergence. It is also known as Maclaurin-Cauchy Test.

Let N be a natural number (non-negative number), and it is a monotonically decreasing function, then the function is defined as

f: [N,∞ ]→ ℝ

Then the series $$\sum_{m=N}^{\infty }f(m)$$ is convergent if and only if the integral $$\int_{N}^{\infty }f(t)dt$$ is finite.

## Conditions for Integral Test

The integral comparison test is mainly for the integral terms. If we have two functions say f(x) and g(x) in such a way that g(x)≥ f(x) on the given interval [c, ∞], then it should have the following conditions.

• If the term $$\int_{c}^{\infty }g(x)dx$$ converges, then the term so does $$\int_{c}^{\infty }f(x)dx$$.
• If the term $$\int_{c}^{\infty }f(x)dx$$ divergences, then the term so does $$\int_{c}^{\infty }g(x)dx$$.

## Integral Test Proof

The integral test proof depends on the comparison test. We know that,

$$\int_{N}^{\infty }f(t)dt$$ is nothing but the sum of the series $$\sum_{m=N}^{\infty }\int_{m}^{m+1}f(t)dt$$

Since “f” is monotonically decreasing function, then

f(t) ≤ f(m) for every “t” in [m, ∞]

For m > N, $$\int_{m}^{m+1}f(t)dt$$ ≤ $$\int_{m}^{m+1}f(m)dt$$ = f(m)

It means that $$\int_{m}^{m+1}f(t)dt$$ ≤ f(m)

As both the quantities are non-negative, use comparison test.

If $$\sum_{m=N}^{\infty }f(m)$$ converges, then $$\sum_{m=N}^{\infty }\int_{m}^{m+1}f(t)dt$$ = $$\int_{N}^{\infty }f(t)dt$$ also converges.

That is, it is finite.

So we are done with one step of the proof.

Now again take that f is monotonically decreasing function, we get

f(m) ≤ f(t) for every “x” in [M, m]

So, f(m) = $$\int_{m-1}^{m}f(t)dt$$ ≤ $$\int_{m-1}^{m}f(t)dt$$

From comparison theorem, we get

If $$\sum_{m=N}^{\infty }\int_{m-1}^{m}f(t)dt = f(N)+\int_{N}^{\infty }f(t)dt$$ converges, then we can say that $$\sum_{m=N}^{\infty }f(m)$$ also converges, which proves the other part of the theorem.

### Integral Test Example with Solution

Question:

Test the convergence of $$\sum_{n=1}^{10}n$$

Solution:

Given: $$\sum_{n=1}^{10}n$$

We can define it as

f: [N,∞ ]→ ℝ

f: [1,10 ]→ ℝ by f(x) = x=n

So, we can write it as

$$\int_{1}^{10}f(x)dx =\int_{1}^{10}xdx$$

= $$\left [ \frac{x^{2}}{2} \right ]_{1}^{10}$$

= (102/2)-(12/2)

= 50 – (½)

= 99/2 = 49.5

Hence, by integral test, the given series is convergent.

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