Section 10.7 : Comparison Test/Limit Comparison Test
7. Determine if the following series converges or diverges.
\[\sum\limits_{n = 0}^\infty {\frac{{{2^n}{{\sin }^2}\left( {5n} \right)}}{{{4^n} + {{\cos }^2}\left( n \right)}}} \]Show All Steps Hide All Steps
Start SolutionFirst, the series terms are,
\[{a_n} = \frac{{{2^n}{{\sin }^2}\left( {5n} \right)}}{{{4^n} + {{\cos }^2}\left( n \right)}}\]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.
The trig functions in the numerator and in the denominator won’t really affect the size of the numerator and denominator for large enough \(n\) and so it seems like for large \(n\) that the term will probably behave like,
\[{b_n} = \frac{{{2^n}}}{{{4^n}}} = {\left( {\frac{2}{4}} \right)^n} = {\left( {\frac{1}{2}} \right)^n}\]We also know that the series,
\[\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{2}} \right)}^n}} \]will converge because it is a geometric series with \(r = \frac{1}{2} < 1\).
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.
In this case we can work with both the numerator and the denominator. Let’s start with the numerator. We know that \(0 \le {\sin ^2}\left( {5n} \right) \le 1\) and so replacing the \({\sin ^2}\left( {5n} \right)\) in the numerator with the largest possible value we get,
\[{2^n}{\sin ^2}\left( {5n} \right) < {2^n}\left( 1 \right) = {2^n}\]Using this we can make the numerator larger to get the following relationship,
\[\frac{{{2^n}{{\sin }^2}\left( {5n} \right)}}{{{4^n} + {{\cos }^2}\left( n \right)}} < \frac{{{2^n}}}{{{4^n} + {{\cos }^2}\left( n \right)}}\]Now, in the denominator we know that \(0 \le {\cos ^2}\left( n \right) \le 1\) and so replacing the \({\cos ^2}\left( n \right)\) with the smallest possible value we get,
\[{4^n} + {\cos ^2}\left( n \right) > {4^n} + 0 = {4^n}\]Using this we can make the denominator smaller (and hence make the rational expression larger) to get,
\[\frac{{{2^n}{{\sin }^2}\left( {5n} \right)}}{{{4^n} + {{\cos }^2}\left( n \right)}} < \frac{{{2^n}}}{{{4^n} + {{\cos }^2}\left( n \right)}} < \frac{{{2^n}}}{{{4^n}}} = {\left( {\frac{1}{2}} \right)^n}\] Show Step 4Now, the series,
\[\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{2}} \right)}^n}} \]is a convergent series (as discussed above) and we’ve also shown that the series terms in this series are larger 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 converge.
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.