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

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

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.

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

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.