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17Calculus Infinite Series - Practice Problem 2436

Infinite Series Practice Problem 2436

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

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. This is not a telescoping series or one of the special series.
2. We can't use the integral test because there are some negative terms and the integral test requires all positive terms.

What We Are Left With

3. If we look closely, the term \(\cos(n\pi) = (-1)^n\), so we have alternating series test. If the alternating series test does not work, we can move on to some others, but we want to try the divergence test and the alternating series test first.
Note - Once we determine that we have an alternating series, we immediately try the alternating series test, since most of the time, it will work. If it doesn't for some reason, we can come back and consider other series tests.

Applying The Tests

Divergence Test

\(\displaystyle{ a_n = \frac{n~\cos(n\pi)}{2n-1} }\)

\(\displaystyle{ \lim_{n \to \infty}{ \frac{n~\cos(n\pi)}{2n-1} } = \lim_{n \to \infty}{ \frac{\cos(n\pi)}{2-1/n} } }\)

In this limit, we know that \(\displaystyle{ \lim_{n \to \infty}{1/n} = 0 }\), leaving \(2\) in the denominator. In the numerator, \(\cos(n\pi)\) oscillates between \(-1\) and \(1\). So we are left with the series oscillating between \(-1/2\) and \(1/2\) and therefore, the limit \(\displaystyle{ \lim_{n \to \infty}{ \frac{n~\cos(n\pi)}{2n-1} } \neq 0 }\)
So, by the divergence test, the series diverges.

Conclusion
The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n~\cos(n\pi)}{2n-1} } }\) diverges by the divergence test.

Alternating Series Test

If you look closely, you will notice that \(\cos(n\pi)\) is just a fancy way of writing \((-1)^{n}\). This is an alternating series which can be written \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(-1)^{n} n}{2n-1} } }\)
Let's try the alternating series test with \(\displaystyle{ a_n = \frac{n}{2n-1} }\)

Condition 1: We need to find the limit of \(\displaystyle{ \lim_{n \to \infty}{a_n} }\).
\(\displaystyle{ \lim_{n \to \infty}{a_n} = \lim_{n \to \infty}{ \frac{n}{2n-1} \frac{1/n}{1/n} } = \lim_{n \to \infty}{ \frac{1}{2-1/n} } = \frac{1}{2} \neq 0 }\)

Since the limit is not zero, condition 1 does not hold, so the alternating series test is inconclusive.

Test Summary List and Answer

test/series

works?

notes

divergence test

yes

best test

p-series

no

not a p-series

geometric series

no

not a geometric series

alternating series test

no

condition 1 does not hold

telescoping series

no

not a telescoping series

ratio test

limit comparison test

direct comparison test

integral test

no

integral test requires positive terms

root test

Final Answer

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

Notes

Once we noticed that we had an alternating series, we started with that; when condition 1 didn't hold, we knew that the divergence test would tell us that the series diverges.

Really UNDERSTAND Calculus

Topics You Need To Understand For This Page

all infinite series topics

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