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### Section 10.7 : Comparison Test/Limit Comparison Test

10. Determine if the following series converges or diverges.

\[\sum\limits_{n = 1}^\infty {\frac{{\sqrt {2{n^2} + 4n + 1} }}{{{n^3} + 9}}} \]Show All Steps Hide All Steps

Start SolutionFirst, the series terms are,

\[{a_n} = \frac{{\sqrt {2{n^2} + 4n + 1} }}{{{n^3} + 9}}\]and it should pretty obvious that for the range of \(n\) we have in this series 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!

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

For large enough \(n\) we know that the \(2{n^2}\) (a quadratic term) in the numerator will increase at a much faster rate than the \(4n + 1\) (a linear term) portion of the numerator. Therefore the \(2{n^2}\) portion of the numerator will, in all likelihood, define the behavior of the numerator. Likewise, the “+9” in the denominator will not affect the size of the denominator for large \(n\) and so the terms should behave like,

\[{b_n} = \frac{{\sqrt {2{n^2}} }}{{{n^3}}} = \frac{{\sqrt 2 }}{{{n^2}}}\]We also know that the series,

\[\sum\limits_{n = 1}^\infty {\frac{{\sqrt 2 }}{{{n^2}}}} \]will converge by the \(p\)-series Test (\(p = 2 > 1\)).

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

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

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

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

We now have a problem however. The obvious thing to try is to drop the last two terms in the numerator and the last term in the denominator. Doing that however gives the following inequalities,

\[2{n^2} < 2{n^2} + 4n + 1\hspace{0.5in}{n^3} + 9 > {n^3}\]This leads to a real problem! If we use the inequality for the numerator we’re going to get a smaller rational expression and if we use the inequality for the denominator we’re going to get a larger rational expression. Because these two can’t both be used at the same time it will make it fairly difficult to use the Comparison Test since neither one individually give a series we can quickly deal with.

So, the Comparison Test won’t easily work in this case. That pretty much leaves the Limit Comparison Test to try. This test only requires positive terms (which we have) and a second series that we’re pretty sure behaves like the series we want to know the convergence for. Note as well that, for the Limit Comparison Test, we don’t care if the terms for the second series are larger or smaller than problem statement series terms.

If you think about it we already have exactly what we need. In Step 2 we used a second series to guess at the convergence of the problem statement series. The terms in the new series are positive (which we need) and we’re pretty sure it behaves in the same manner as the problem statement series.

So, let’s compute the limit we need for the Limit Comparison Test.

\[\begin{align*}c & = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \left[ {{a_n}\frac{1}{{{b_n}}}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{\sqrt {2{n^2} + 4n + 1} }}{{{n^3} + 9}}\,\,\frac{{{n^2}}}{{\sqrt 2 }}} \right]\\ & = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^2}\sqrt {{n^2}\left( {2 + \frac{4}{n} + \frac{1}{{{n^2}}}} \right)} }}{{\sqrt 2 \,{n^3}\left( {1 + \frac{9}{{{n^3}}}} \right)}}} \right] = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^2}\left( n \right)\sqrt {2 + \frac{4}{n} + \frac{1}{{{n^2}}}} }}{{\sqrt 2 \,{n^3}\left( {1 + \frac{9}{{{n^3}}}} \right)}}} \right] = \frac{{\sqrt 2 }}{{\sqrt 2 }} = 1\end{align*}\] Show Step 5Okay. We now have \(0 < c = 1 < \infty \), *i.e.* \(c\) is not zero or infinity and so by the Limit Comparison Test the two series must have the same convergence. We determined in Step 2 that the second series converges and so the series given in the problem statement must also **converge**.

Be careful with the Comparison Test. Too often students just try to “force” larger or smaller by just hoping that the second series terms has the correct relationship (*i.e.* larger or smaller as needed) to the problem series terms. The problem is that this often leads to an incorrect answer. Be careful to always prove the larger/smaller nature of the series terms and if you can’t get a series term of the correct larger/smaller nature then you may need to resort to the Limit Comparison Test.