Integral Test

Calculus 2

Integral Test

Let {an}n=1\left\{ a_n \right\}_{n=1}^\infty be a sequence of positive terms. Suppose there is a positive integer NN such that for all nNn \geq N, an=f(n)a_n = f(n), where f(x)f(x) is a:

  • Decreasing,
  • Continuous,
  • Positive,

function of xx. Then, the series n=Nan\sum_{n=N}^\infty a_n and the integral Nf(x)dx\int_N^\infty f(x) \, dx both converge or both diverge.

In other words, the Integral Test states that if f(x)f(x) is the continuous function corresponding to ana_n and f(x)f(x) is decreasing, continuous, and positive, then n=Nan\sum_{n=N}^\infty a_n and Nf(x)dx\int_N^\infty f(x) \, dx either both converge or both diverge.

Keep in mind the following:

  • If the series and integral are convergent, they are not necessarily equal.
  • The integral need only be decreasing eventually on the interval.
  • The Integral Test can be used to prove the conclusion of the p-Series Test.

Example 1

Determine the convergence/divergence of the series n=11n+2\sum_{n=1}^\infty \frac{1}{n+2} using the Integral Test.

View Answer

Diverges

Example 2

Determine the convergence/divergence of the series n=18nn2+4\sum_{n=1}^\infty \frac{8n}{n^2+4} using the Integral Test.

View Answer

Diverges

Example 3

Determine the convergence/divergence of the series n=110n2+1\sum_{n=1}^\infty \frac{-10}{n^2+1} using the Integral Test.

View Answer

Inconclusive (terms are not positive)

Example 4

Determine the convergence/divergence of the series n=1nen2\sum_{n=1}^\infty n e^{-n^2} using the Integral Test.

View Answer

Converges

Example 5

Show that the p-Series n=11np\sum_{n=1}^\infty \frac{1}{n^p} converges for p>1p > 1 and diverges for p1p \leq 1 using the Integral Test (assuming pp is a positive real constant).

View Answer

Evaluate 11xpdx\int_1^\infty \frac{1}{x^p} \, dx:

  • For p>1p > 1, the integral converges.
  • For p1p \leq 1, the integral diverges.

Thus, the series follows the same behavior as the integral.


Remainder Estimation

For some convergent series, such as geometric series or telescoping series, we can find the sum of the series. For most convergent series, however, we cannot easily find the sum. Nevertheless, we can estimate the sum by adding the first nn terms to get the partial sum SnS_n, but we need to know how far off SnS_n is from the total sum SS.

If a series an\sum a_n is known to be convergent by the Integral Test, the remainder RnR_n measuring the difference between the total sum SS and its nthn^\text{th} partial sum SnS_n is given by:

Rn=SSn=an+1+an+2+an+3+.R_n = S - S_n = a_{n+1} + a_{n+2} + a_{n+3} + \cdots.

If {an}n=1\{a_n\}_{n=1}^\infty is a sequence of positive terms with ak=f(k)a_k = f(k), where f(x)f(x) is a continuous, positive, decreasing function for xnx \geq n and k=1ak\sum_{k=1}^\infty a_k converges to SS, then the remainder Rn=SSnR_n = |S - S_n| satisfies:

n+1f(x)dxRnnf(x)dx.\int_{n+1}^\infty f(x) \, dx \leq R_n \leq \int_n^\infty f(x) \, dx.

Example 6

Estimate the sum SS of the series n=11n2\sum_{n=1}^\infty \frac{1}{n^2} with n=10n = 10.


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