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Integral Test

Sequences and series act as a building block for the analysis process and the continuity of the function can be easily proved using the sequences. The integral test is used to find whether the given series is converged 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 convergent while dealing with some specific functions on sequences and series. In this article, you will learn about the procedure of integral test, Integral Test of Convergence proof and comparison tests.

Learn: Series

Integral Test for Convergence

The integral test for convergence is used to test the infinite series of non-negative terms for convergence, and it is also known as the 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

\(\begin{array}{l}\sum_{m=N}^{\infty }f(m)\end{array} \)
is convergent if and only if the integral
\(\begin{array}{l}\int_{N}^{\infty }f(t)dt\end{array} \)
is finite.

Learn about Increasing and decreasing functions and monotonicity in detail.

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

Integral Test Proof

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

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

Since “f” is monotonically decreasing function, then

f(t) ≤ f(m) for every “t” in [m, ∞]

For m > N,

\(\begin{array}{l}\int_{m}^{m+1}f(t)dt\end{array} \)
\(\begin{array}{l}\int_{m}^{m+1}f(m)dt\end{array} \)
= f(m)

It means that

\(\begin{array}{l}\int_{m}^{m+1}f(t)dt\end{array} \)
≤ f(m)

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

If

\(\begin{array}{l}\sum_{m=N}^{\infty }f(m)\end{array} \)
converges, then
\(\begin{array}{l}\sum_{m=N}^{\infty }\int_{m}^{m+1}f(t)dt\end{array} \)
=
\(\begin{array}{l}\int_{N}^{\infty }f(t)dt\end{array} \)
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) =

\(\begin{array}{l}\int_{m-1}^{m}f(t)dt\end{array} \)
\(\begin{array}{l}\int_{m-1}^{m}f(t)dt\end{array} \)

From comparison theorem, we get

If

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

Integral Test Examples with Solutions

Example 1:

Test the convergence of

\(\begin{array}{l}\sum_{n=1}^{10}n\end{array} \)

Solution:

Given:

\(\begin{array}{l}\sum_{n=1}^{10}n\end{array} \)

We can define it as

f: [N,∞ ]→ ℝ

f: [1,10 ]→ ℝ by f(x) = x=n

So, we can write it as

\(\begin{array}{l}\int_{1}^{10}f(x)dx =\int_{1}^{10}xdx\end{array} \)

=

\(\begin{array}{l}\left [ \frac{x^{2}}{2} \right ]_{1}^{10}\end{array} \)

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

= 50 – (½)

= 99/2 = 49.5

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

Let’s have a look at another example of the Integral Test for Infinite Series.

Example 2:

Using the integral test, determine whether the series

\(\begin{array}{l}\sum_{n=1}^{\infty}\frac{1}{2n+5}\end{array} \)
is convergent.

Solution:

Given,

\(\begin{array}{l}\sum_{n=1}^{\infty}\frac{1}{2n+5}\end{array} \)

Let us take the function f(x) = 1/(2x + 5)

f'(x) = -2/(2x + 5)2 < 0

Thus, f(x) is positive and continuous on the interval [1, ∞) and it is a decreasing function.

f(n) = 1/(2n + 5)

As per the integral test, to check whether the given series is convergent, we need to estimate

\(\begin{array}{l}\int_{1}^{\infty}f(x) dx\end{array} \)
.

So, 

\(\begin{array}{l}\int_{1}^{\infty}\frac{1}{2x+5} dx\end{array} \)

This can also be written as:

\(\begin{array}{l}\lim_{t\rightarrow \infty}\int_{1}^{t}\frac{1}{2x+5} dx\end{array} \)

Now, integrating the function, we get;

\(\begin{array}{l}=\lim_{t\rightarrow \infty}\left [ \frac{1}{2}ln(2x+5) \right ]_{1}^{t}\end{array} \)

Apply the limits,

\(\begin{array}{l}= \frac{1}{2}\lim_{t\rightarrow \infty}\left [ln(2t+5) -ln(7)\right ]_{1}^{t}=\infty\end{array} \)

Therefore, the given series is not convergent and diverges to infinity.

We can observe several examples of series while determining the convergence. Suppose a series of the form

\(\begin{array}{l}\sum_{n=1}^{\infty}\frac{1}{n^p}=1+ \frac{1}{2^p}+\frac{1}{3^p}+\frac{1}{4^p}+…+\frac{1}{n^p}+…\end{array} \)
called p-series, where p is a constant such that the series is converges is p > 1 and diverges is p ≤ 1.

Integral Test Problems

  1. Use the integral test to check whether the series
    \(\begin{array}{l}\sum_{n=1}^{\infty}\frac{4}{n^2}\end{array} \)
    converges.
  2. Test the convergence of the series
    \(\begin{array}{l}\sum_{n=1}^{\infty}\frac{3}{\sqrt{n}}\end{array} \)
    using an integral test.
  3. Check whether the series
    \(\begin{array}{l}\sum_{n=1}^{\infty}ne^{-2n}\end{array} \)
    is convergent.

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