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17Calculus Infinite Series - Practice Problem 2435

Infinite Series Practice Problem 2435


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 an alternating series, telescoping series or one of the special series. Although there are some positive and negative terms, if you write out the first few terms, you can tell that they are not alternating or telescoping.
2. Also, a quick calculation tells us that the limit of \(a_n\) is zero. So the divergence test will not help us.
3. The form of \(a_n\) contains \(\sin(n)\) in the numerator and a polynomial in the denominator. This combination is a problem to integrate. So we set aside the use of the integral test for the moment.
4. Applying the root test will give us a complicated denominator and \(\sqrt[n]{\sin(n)}\) in the numerator. So we will put that aside for now and come back to it only if nothing else works.

What We Are Left With

5. Now we are left with the ratio test and the two comparison tests, the limit comparison test and the direct comparison test. There is nothing else that jumps out to tell us that one of them will work best. So we will try them in order as they appear in the infinite series table.

Applying The Tests

Ratio Test

\(\displaystyle{ a_n = \frac{\sin(n)}{n^3+n+1} }\) and \(\displaystyle{ a_{n+1} = \frac{\sin(n+1)}{(n+1)^3+(n+1)+1} }\)

\( \begin{array}{rcl} \displaystyle{\lim_{n \to \infty}{ \abs{\frac{a_{n+1}}{a_n}} }} & = & \displaystyle{\lim_{n \to \infty}{ \abs{\frac{\sin(n+1)}{(n+1)^3+(n+1)+1} \frac{n^3+n+1}{\sin(n)}} } } \\ & = & \displaystyle{\lim_{n \to \infty}{ \abs{\frac{\sin(n+1)}{\sin(n)} \frac{n^3+n+1}{(n+1)^3+(n+1)+1}} } } \end{array} \)

Unfortunately, we have a term \(\sin(n+1)/\sin(n)\) whose limit cannot be evaluated. To see this, we do a little bit of trig.

\(\sin(n+1) = \sin(n)\cos(1)+\cos(n)\sin(1)\)

\(\displaystyle{ \frac{\sin(n)\cos(1)+\cos(n)\sin(1)}{\sin(n)} = \cos(1) + \sin(1)\frac{\cos(n)}{\sin(n)} = \cos(1)+ \sin(1)\cot(n) }\)

Looking at the graph of \(y=\cot(x)\), you can see that it oscillates. So the limit does not exist. This means the ratio test is inconclusive.

Since the limit \(\displaystyle{ \lim_{n \to \infty}{ \abs{\frac{a_{n+1}}{a_n}} } }\) does not exist, the ratio test is inconclusive.

Limit Comparison Test

\(\displaystyle{ a_n = \frac{\abs{\sin(n)}}{n^3+n+1} }\)
For large \(n\) the \(n^3\) term dominates in the denominator and \(\abs{\sin(n)} \leq 1 \). So let's compare this series with the test series \(\sum{t_n}\) where \(\displaystyle{t_n = \frac{1}{n^3} }\).
Since \(t_n\) is a p-series with \(p=3\), the test series converges. The limit comparison test requires us to set up the limit \(\displaystyle{ \lim_{n \to \infty}{\frac{a_n}{t_n}} }\).
\( \begin{array}{rcl} \displaystyle{\lim_{n \to \infty}{\frac{a_n}{t_n}}} & = & \displaystyle{\lim_{n \to \infty}{ \frac{\abs{\sin(n)}}{n^3+n+1} \frac{n^3}{1} }} \ & = & \displaystyle{\lim_{n \to \infty}{ \frac{\abs{\sin(n)}}{1+1/n^2+1/n^3} }} \end{array} \)
This last limit does not exist since the limit of the numerator does not exist. So the limit comparison test cannot be used to prove convergence or divergence.

Direct Comparison Test

\(\displaystyle{ a_n = \frac{\sin(n)}{n^3+n+1} }\)

Before we start with the direct comparison test, we have to look carefully at the requirements. Notice that in both inequalities, we require that \(a_n > 0 \). However, for this series, the sine term introduces some negative terms. So we can't just directly use the direct comparison test. This is an important detail. If we don't handle this detail, we can't use the direct comparison test.

Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if \(\sum{\abs{a_n}}\) converges, then \(\sum{a_n}\) also converges. In this case,
\(\displaystyle{ \abs{a_n} = \abs{\frac{\sin(n)}{n^3+n+1}} = \frac{\abs{\sin(n)}}{\abs{n^3+n+1}} }\)
Since \(n > 0\) the denominator is always positive so we just need to replace \(\sin(n)\) with \(\abs{\sin(n)}\). So, in this problem, we will work toward convergence. If we get the result that the series diverges, then we can't use the absolute convergence theorem and the result is inconclusive.

For large \(n\) the \(n^3\) term dominates. In the numerator, \(\abs{\sin(n)}\) is always less than or equal to one, let's compare this series with the test series \(\sum{t_n}\) where \(\displaystyle{t_n = \frac{1}{n^3} }\).

Since \(t_n\) is a p-series with \(p=3>1\), the test series converges. So the direct comparison test requires us to set up the inequality as \(a_n \leq t_n\).

\(\begin{array}{rcl} a_n & \leq & t_n \\ \displaystyle{\frac{\abs{\sin(n)}}{n^3+n+1}} & \leq & \displaystyle{\frac{1}{n^3}} \\ \abs{\sin(n)} & \leq & \displaystyle{\frac{n^3+n+1}{n^3}} \\ \abs{\sin(n)} & \leq & \displaystyle{1 + \frac{1}{n^2} + \frac{1}{n^3}} \end{array}\)

The last inequality is always true since \(\abs{\sin(n)}\leq 1\) and the right side is always greater than one. So the series converges.

The series \(\sum{a_n}\) converges by the direct comparison test and converges absolutely by the absolute convergence theorem.
There is a similar practice problem on the direct comparison test page.

Test Summary List and Answer




divergence test


limit is zero, so test is inconclusive



not a p-series

geometric series


not a geometric series

alternating series test


not an alternating series

telescoping series


not a telescoping series

ratio test



limit comparison test


unable to evaluate limit

direct comparison test


integral test


not integrable

root test


too complicated

Final Answer

The series converges absolutely.


1. To be able to use the direct comparison test, we had to use the absolute convergence theorem.
2. This is one of the few series problems that can be easily solved with only one test. Most series convergence/divergence can be proved using more than one test.

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