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Section 10.7 : Comparison Test/Limit Comparison Test

1. Determine if the following series converges or diverges.

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

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First, the series terms are,

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

and it should pretty obvious in this case that they are positive and so we know that we can use the Comparison Test on 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!

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For most of the Comparison Test problems we usually guess the convergence and proceed from there. However, in this case it is hopefully clear that for any \(n\),

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

Now, let’s take a look at the following series,

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

Because \(\mathop {\lim }\limits_{n \to \infty } 1 = 1 \ne 0\) we can see from the Divergence Test that this series will be divergent.

So we’ve found a divergent series with terms that are smaller than the original series terms. Therefore, by the Comparison Test the series in the problem statement must also be divergent.

As a final note for this problem notice that we didn’t actually need to do a Comparison Test to arrive at this answer. We could have just used the Divergence Test from the beginning since,

\[\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{1}{{{n^2}}} + 1} \right)^2} = 1 \ne 0\]

This is something that you should always keep in mind with series convergence problems. The Divergence Test is a quick test that can, on occasion, be used to quickly determine that a series diverges and hence avoid a lot of the hassles of some of the other tests.