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Section 10.6 : Integral Test

3. Determine if the following series converges or diverges.

\[\sum\limits_{n = 2}^\infty {\frac{1}{{{{\left( {2n + 7} \right)}^3}}}} \]

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Start Solution

Okay, prior to using the Integral Test on this series we first need to verify that we can in fact use the Integral Test!

Show Step 2

The series terms are,

\[{a_n} = \frac{1}{{{{\left( {2n + 7} \right)}^3}}}\]

We can clearly see that for the range of \(n\) in the series the terms are positive and so that condition is met.

Show Step 3

In this case because there is only one \(n\) in the denominator and because all the terms in the denominator are positive it is (hopefully) clear that,

\[{a_n} = \frac{1}{{{{\left( {2n + 7} \right)}^3}}} > \frac{1}{{{{\left( {2\left( {n + 1} \right) + 7} \right)}^3}}} = {a_{n + 1}}\]

and so the series terms are decreasing.

Okay, we now know that both of the conditions required for us to use the Integral Test have been verified we can proceed with the Integral Test.

It is very important to always check the conditions for a particular series test prior to actually using the test. One of the biggest mistakes that many students make with the series test is using a test on a series that don’t meet the conditions for the test and getting the wrong answer because of that!

Show Step 4

Now, let’s compute the integral for the test.

\[\begin{align*}\int_{2}^{\infty }{{\frac{1}{{{{\left( {2x + 7} \right)}^3}}}\,dx}} & = \mathop {\lim }\limits_{t \to \infty } \int_{2}^{t}{{\frac{1}{{{{\left( {2x + 7} \right)}^3}}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left( { - \frac{1}{4}\frac{1}{{{{\left( {2x + 7} \right)}^2}}}} \right)} \right|_2^t\\ & = \mathop {\lim }\limits_{t \to \infty } \left( { - \frac{1}{4}\frac{1}{{{{\left( {2t + 7} \right)}^2}}} + \frac{1}{4}\frac{1}{{{{\left( {11} \right)}^2}}}} \right) = \frac{1}{{484}}\end{align*}\] Show Step 5

Okay, the integral from the last step is a convergent integral and so by the Integral Test the series must also be a convergent series.