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

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

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

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

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.

Conclusion
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.