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Section 10.7 : Comparison Test/Limit Comparison Test

6. Determine if the following series converges or diverges.

\[\sum\limits_{n = 1}^\infty {\frac{{2{n^3} + 7}}{{{n^4}{{\sin }^2}\left( n \right)}}} \]

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Start Solution

First, the series terms are,

\[{a_n} = \frac{{2{n^3} + 7}}{{{n^4}{{\sin }^2}\left( n \right)}}\]

and it should pretty obvious that they are positive and so we know that we can attempt the Comparison Test for this series.

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!

Hint : Can you make a guess as to whether or not the series should converge or diverge?
Show Step 2

Let’s first see if we can make a reasonable guess as to whether this series converges or diverges.

The “+7” in the numerator and the “\({\sin ^2}\left( n \right)\)” in the denominator won’t really affect the size of the numerator and denominator respectively for large enough \(n\) and so it seems like for large \(n\) that the term will probably behave like,

\[{b_n} = \frac{{2{n^3}}}{{{n^4}}} = \frac{2}{n}\]

We also know that the series,

\[\sum\limits_{n = 1}^\infty {\frac{2}{n}} \]

will diverge because it is a harmonic series or by the \(p\)-series Test.

Therefore, it makes some sense that we can guess the series in the problem statement will probably diverge as well.

Hint : Now that we have our guess, if we’re going to use the Comparison Test, do we need to find a series with larger or a smaller terms that has the same convergence/divergence?
Show Step 3

So, because we’re guessing that the series diverges we’ll need to find a series with smaller terms that we know, or can prove, diverges.

Note as well that we’ll also need to prove that the new series terms really are smaller than the terms from the series in the problem statement. We can’t just “hope” that the will be smaller.

In this case, because the terms in the problem statement series are a rational expression, we know that we can make the series terms smaller by either making the numerator smaller or the denominator larger.

In this case we can work with both the numerator and the denominator. Let’s start with the numerator. It should be pretty clear that,

\[2{n^3} < 2{n^3} + 7\]

Using this we can make the numerator smaller to get the following relationship,

\[\frac{{2{n^3} + 7}}{{{n^4}{{\sin }^2}\left( n \right)}} > \frac{{2{n^3}}}{{{n^4}{{\sin }^2}\left( n \right)}}\]

Now, we know that \(0 \le {\sin ^2}\left( n \right) \le 1\) and so in the denominator we can see that if we replace the \({\sin ^2}\left( n \right)\) with its largest possible value we have,

\[{n^4}{\sin ^2}\left( n \right) < {n^4}\left( 1 \right) = {n^4}\]

Using this we can make the denominator larger (and hence make the rational expression smaller) to get,

\[\frac{{2{n^3} + 7}}{{{n^4}{{\sin }^2}\left( n \right)}} > \frac{{2{n^3}}}{{{n^4}{{\sin }^2}\left( n \right)}} > \frac{{2{n^3}}}{{{n^4}}} = \frac{2}{n}\] Show Step 4

Now, the series,

\[\sum\limits_{n = 1}^\infty {\frac{2}{n}} \]

is a divergent series (as discussed above) and we’ve also shown that the series terms in this series are smaller than the series terms from the series in the problem statement.

Therefore, by the Comparison Test the series given in the problem statement must also diverge.

Be careful with these kinds of problems. The series we used in Step 2 to make the guess ended up being the same series we used in the Comparison Test and this will often be the case but it will not always be that way. On occasion the two series will be different.