## 17Calculus Infinite Series - Practice Problem 2438

Infinite Series Practice Problem 2438

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2-\cos^2(n)}}}$$

Determine the convergence or divergence of the series.
If the series converges,
- determine the value the series converges to, if possible; and
- determine if the series converges absolutely or conditionally.

Choosing Which Tests To Try

Rejecting the Obvious

1. We can tell right away that this is not an alternating series, telescoping series or one of the special series.
2. We can tell that the limit of $$a_n$$ is zero. So the divergence test will not help us.
3. The form of $$a_n$$ contains $$\cos^2(n)$$ in the denominator and a polynomial in the denominator. This combination is a problem to integrate.

What We Are Left With

4. We are left with the ratio test, the comparison tests and the root test. The root test will give us quite a complicated term in the denominator. So we will not try that one unless nothing else works.

Applying The Tests

### Ratio Test

$$\displaystyle{ a_n = \frac{n}{n^2-\cos^2(n)} }$$ and $$\displaystyle{ a_{n+1} = \frac{n+1}{(n+1)^2-\cos^2(n+1)} }$$

$$\displaystyle{ \begin{array}{rcl} \displaystyle{\lim_{n \to \infty}{ \left| \frac{a_{n+1}}{a_n} \right| }} & = & \displaystyle{\lim_{n \to \infty}{ \left| \frac{n+1}{(n+1)^2-\cos^2(n+1)} \frac{n^2-\cos^2(n)}{n} \right| }} \\ & = & \displaystyle{\lim_{n \to \infty}{ \left| \frac{n+1}{n} \frac{n^2-\cos^2(n)}{(n+1)^2-\cos^2(n+1)} \right| }} \\ & = & \displaystyle{\lim_{n \to \infty}{ \left| \left[ 1 + \frac{1}{n} \right] \frac{n^2-\cos^2(n)}{(n+1)^2-\cos^2(n+1)}\frac{1/n^2}{1/n^2} \right| }} \\ & = & \displaystyle{\lim_{n \to \infty}{ \left| \left[ 1 + \frac{1}{n} \right] \frac{1-\cos^2(n)/n^2}{(1+1/n)^2-\cos^2(n+1)/n^2} \right| } } \end{array} }$$

Now let's look at each piece.
$$\displaystyle{ \lim_{n \to \infty}{ 1/n } = 0 }$$

Using the pinching theorem, we know that $$\displaystyle{ \lim_{n \to \infty}{ \frac{\cos^2(n)}{n^2} } =0 }$$ and $$\displaystyle{ \lim_{n \to \infty}{ \frac{\cos^2(n+1)}{n^2} } =0 }$$. The details are in the section below.

So now we have $$\displaystyle{ \lim_{n \to \infty}{ \left| \left[ 1 + \frac{1}{n} \right] \frac{1-\cos^2(n)/n^2}{(1+1/n)^2-\cos^2(n+1)/n^2} \right| } = (1+0)\frac{1-0}{1-0} = 1 }$$

Conclusion
Since the limit $$\displaystyle{ \lim_{n \to \infty}{ \left| \frac{a_{n+1}}{a_n} \right| } = 1}$$ the ratio test is inconclusive.

### Limit Comparison Test

$$\displaystyle{ a_n = \frac{n}{n^2-\cos^2(n)} }$$

For large $$n$$ the $$n^2$$ term dominates in the denominator since $$\cos^2(n) \leq 1$$. In the numerator, we have $$n$$. So let's compare this series with the test series $$\sum{t_n}$$ where $$\displaystyle{t_n = \frac{1}{n} }$$.

Since $$t_n$$ is a p-series with $$p=1$$, the test series diverges. The limit comparison test requires us to set up the limit $$\displaystyle{ \lim_{n \to \infty}{\frac{a_n}{t_n}} }$$.

$$\begin{array}{rcl} \displaystyle{\lim_{n \to \infty}{\frac{a_n}{t_n}}} & = & \displaystyle{\lim_{n \to \infty}{\frac{n}{n^2-\cos^2(n)} \frac{n}{1} }} \\ & = & \displaystyle{\lim_{n \to \infty}{\frac{n^2}{n^2-\cos^2(n)} \frac{1/n^2}{1/n^2} }} \\ & = & \displaystyle{\lim_{n \to \infty}{\frac{1}{1-\cos^2(n)/n^2} }} \\ & = & \displaystyle{\frac{1}{1-0}} = 1 \end{array}$$
Since the limit is finite and positive and the test series diverges, the series $$\sum{a_n}$$ also diverges.

Conclusion
The series $$\sum{a_n}$$ diverges by the limit comparison test.

We evaluated the limit $$\displaystyle{ \lim_{n \to \infty}{\frac{\cos^2(n)}{n^2}} }$$ using the pinching theorem. The details can be found in the section below.

### Direct Comparison Test

$$\displaystyle{ a_n = \frac{n}{n^2-\cos^2(n)} }$$

For large $$n$$ the $$n^2$$ term dominates in the denominator since $$\cos^2(n) \leq 1$$. In the numerator, we have $$n$$. So let's compare this series with the test series $$\sum{t_n}$$ where $$\displaystyle{t_n = \frac{1}{n} }$$.

Since $$t_n$$ is a p-series with $$p=1$$, the test series diverges. So the direct comparison test requires us to set up the inequality as $$t_n \leq a_n$$.

$$\begin{array}{rcl} t_n & \leq & a_n \\ \displaystyle{\frac{1}{n}} & \leq & \displaystyle{\frac{n}{n^2 - \cos^2(n)}} \\ n^2 - \cos^2(n) & \leq & n^2 \\ -\cos^2(n) & \leq & 0 \\ \cos^2(n) & \geq & 0 \end{array}$$

The last inequality is always true since a squared term is always greater than or equal to zero. So the series diverges.

Conclusion
The series $$\sum{a_n}$$ diverges by the direct comparison test.

Evaluating Limit With The Pinching Theorem

$$\begin{array}{rcccl} 0 & \leq & \displaystyle{\frac{\cos^2(n)}{n^2}} & \leq & \displaystyle{\frac{1}{n^2}} \\ \displaystyle{\lim_{n \to \infty}{0}} & \leq & \displaystyle{\lim_{n \to \infty}{\frac{\cos^2(n)}{n^2} } } & \leq & \displaystyle{\lim_{n \to \infty}{\frac{1}{n^2}}} \\ 0 & \leq & \displaystyle{\lim_{n \to \infty}{\frac{\cos^2(n)}{n^2} }} & \leq & 0 \end{array}$$
By the pinching theorem, $$\displaystyle{ \lim_{n \to \infty}{\frac{\cos^2(n)}{n^2} } = 0 }$$
Using the same logic, we can say that $$\displaystyle{ \lim_{n \to \infty}{\frac{\cos^2(n+1)}{n^2} } = 0 }$$

Test Summary List and Answer

test/series works? notes no limit is zero, so test is inconclusive no not a p-series no not a geometric series no not an alternating series no not a telescoping series no inconclusive yes yes best test no not integrable no too complicated

The series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2-\cos^2(n)}}}$$ diverges.

Notes

The direct comparison test is probably the best test to use here since we did not have to evaluate a limit that involved using the pinching theorem. However, the limit comparison test worked well too.

Really UNDERSTAND Calculus

### Topics You Need To Understand For This Page

 all infinite series topics L'Hopitals Rule

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

free ideas to save on bags & supplies

As an Amazon Associate I earn from qualifying purchases.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

We use cookies on this site to enhance your learning experience.