## Integral Test: Definition

The** Integral Test **is used to prove whether a sequence a_{n} or its corresponding function *f(x)* converges or not; It is defined by approaching a given series of partial sums through integration.

## Integral Test Formulation

The test begins by finding a function *f(n)* that, for the half-closed interval [1, ∞), contains three properties:

- A continuous function,
- All positive values,
- The function is decreasing.

Equating *f(n) = *a_{n} for all *n **≥ **N, *where both *n* and *N* are positive integers, forces both the summation of a_{n} and the integral of *f(x)* to either converge or diverge.

The** integral** needs **to be evaluated first** to have the sequence in question follow suit in the determination of convergence or divergence. If the sequence a_{n} has its representing function *f(x)* decrease in value as the step size *n* increases, then—as n goes to infinity—the y-value for *f(x)* equals zero. Therefore, the sum of all y-values from *n* = 1 to *n* = *N* will conclude the convergence property of a_{n}.

The integral test is **only used** to determine if the given sequence a_{n} converges, but **never** for the sequence’s converging value. There isn’t a process that can evaluate a given series for its exact converging value.

The reasoning is that the integral itself captures the area underneath its curve while the summation of a sequence takes the sum of y-values produced by *f(x)* for each iteration of *n*. This issue is solved through the Remainder Estimation in addition to the logical solving procedure below.

## Integral Test Remainder

For *f(x)* that contains the properties of being continuous, positive, and decreasing for *x* *≥ n _{o} *and for the sequence a

_{n}to being able to converge, an Upper Bound and Lower Bound exists that act as a remainder. The remainder value represents the error when approximating the infinite summation of a

_{n}to the nth partial sum. Since the integral test can’t determine the exact value for the summation of a

_{n},

**there will always be a positive non-zero remainder when performing an approximation attempt.**

Limitations on the Upper and Lower Bounds for the value of the sequence a_{n} can be represented as:

Where:

- s
_{n – 1}represents the summation for a_{n}for one term less than what*n*was set for originally and - s
_{n}represents the summation for a_{n}for*n*terms.

The first inequality group can be represented by two visuals. Graph (a) has the function *f(x)* being integrated above the rectangles that the summation for a_{n} is accounting for. Graph (b) has the same function *f(x)* being integrated below the rectangles that the summation for a_{n} is accounting for (UT Calculus).

The second inequality group represents the underestimation that comes from s_{n} alone and can be added to the integral of *f(x)* to obtain the smallest possible converging value for the sequence a_{n}.

## Estimation Example

A typical series to evaluate can be done for a p-series: a_{n} = n^{-2}. We would like to approximate the sequence’s converging value to three decimal places.

__Step 1__: Set up the summation to then evaluate the integral within the Integral Test.

For an approximation of three decimal places, we limit the integral to have a ceiling of 1/1000 in error.

__Step 2__: Attempt a value N to determine if N needs to be increased to reduce the error of concluding two approximations.

For N^{-1} < 1/1000, the first attempt for a N value can be N = 1000

Then the lower approximation equates to:

And the upper approximation equates to:

The current conclusion rests between the sum equaling to either 1.644 or 1.645. Thus, N needs to be larger to have the sum settling on a single number with three decimal places.

__Step 3__: If the requested decimal rounding rests between two values, increase the *N *value to reevaluate the series converging approximation.

Since the difference in error between the two approximations is *±* 0.0005, we can focus on finding N where N^{-1} = 0.00025. In this case, N = 4000 and the two approximations can be reevaluated.

The new lower approximation equates to:

And the new upper approximation equates to:

Both values, rounded to three decimals, equate to the same value of 1.645.

## Reference

“Estimates of Value of the Series.” *UT Calculus*, web.ma.utexas.edu/users/m408s/m408d/CurrentWeb/LM11-3-5.php.

**CITE THIS AS:**

**Stephanie Glen**. "Integral Test / Remainder Estimate" From

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