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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

divergence test

no

limit is zero, so test is inconclusive

p-series

no

not a p-series

geometric series

no

not a geometric series

alternating series test

no

not an alternating series

telescoping series

no

not a telescoping series

ratio test

no

inconclusive

limit comparison test

yes

direct comparison test

yes

best test

integral test

no

not integrable

root test

no

too complicated

Final Answer

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.

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