Use the test series \(\sum{1/n}\), a divergent p-series.
\(\begin{array}{rcl} \displaystyle{ \lim_{n\to\infty}{ \left| \frac{a_n}{t_n} \right| } } & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^2-1}{n^3+4} \frac{n}{1} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^3-n}{n^3+4} \frac{1/n^3}{1/n^3} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{1-1/n^2}{1+4/n^3} \right] } = 1} \\ \end{array}\)
Since the limit is finite and positive, both series either converge or diverge. Since the test series diverges, so does the original series by the limit comparison test.

For this question \(\displaystyle{ a_n = \frac{n^2-1}{n^3+4} }\). If we think about what happens when n is very large, the \(n^2\) term dominates the 1 in the numerator and the \(n^3\) term dominates the 4 in the denominator. When that happens we are very close to \(n^2/n^3 = 1/n\). So let's apply the direct comparision test using \(t_n = 1/n\). Since \(\sum{1/n}\) is a divergent p-series, we will assume the series \(\sum{a_n}\) diverges also. Using the assumption, the inequality we need to prove is \( 0 \lt t_n \leq a_n \).
It is obvious that \(0 \lt t_n\), so we will focus on showing that \(t_n \leq a_n\).
\(\begin{array}{rcl} t_n & \leq & a_n \\ \displaystyle{ \frac{1}{n} } & \leq & \displaystyle{ \frac{n^2-1}{n^3+4} } \\ n^3+4 & \leq & n(n^2-1) \\ n^3+4 & \leq & n^3-n \\ 4 & \leq & -n \\ -4 & \geq & n \end{array}\)
This last inequality does not hold. So we need another test series.
At this point, we will still assume that the series diverges. So we need a divergent series to compare it to. To find one, we plotted the functions \(f(x)=(x^2-1)/(x^3+4)\) and \(g(x)=1/x\) on the same set of axes. Sure enough, g(x) is larger than f(x) which confirms our work above. Our test series needs to be less than or equal to our original series. So we need a divergent series that is smaller than \(1/n\).
To make \(1/n\) smaller, we can make the denominator larger, so we looked at \(1/(n+1)\). Plotting \(h(x)=1/(x+1)\) on the same set of axes as f(x) and g(x), we can zoom in and see that, sure enough, h(x) is smaller than f(x). But is \(\sum{1/(n+1)}\) still a divergent series? We can use the limit comparison test as follows to find out.
\(\begin{array}{rcl} \displaystyle{ \lim_{n\to\infty}{\frac{t_n}{a_n}} } & = & \displaystyle{ \lim_{n\to\infty}{ \frac{1}{n+1} \frac{n}{1} } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \frac{n}{n+1} } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \frac{1}{1+1/n} } = 1 } \end{array}\)
Since the limit is finite and positive, then both series either converge or diverge. Since the test series diverges, so does \(\sum{1/(n+1)}\).
So now for the direct comparison test, we have a divergent test series \(t_n=1/(n+1)\) and we need to show that \( 0 \lt t_n \leq a_n \). As before \(0 \lt t_n\), so we focus on \(t_n \leq a_n\).
\(\begin{array}{rcl} t_n & \leq & a_n \\ \displaystyle{ \frac{1}{n+1} } & \leq & \displaystyle{ \frac{n^2-1}{n^3+4} } \\ n^3+4 & \leq & (n+1)(n^2-1) \\ n^3+4 & \leq & n^3-n+n^2-1 \\ 4 & \leq & n^2-n-1 \\ 0 & \geq & n^2-n-5 \end{array}\)
Can we find an N such that for \(n \gt N\), \(n^2-n-5 \geq 0\)? Yes we can.
Notice that \(y=x^2-x-5\) is a parabola opening up. So we need to find where the graph crosses the x-axis and choose N larger than that. Completing the square gives us \(x=1/2\pm\sqrt{21/4}\). We don't care about the negative value. The positive value is about 2.79. So we choose N=3. So the inequality \( 0 \geq n^2-n-5 \) holds for \( n \gt 3 \) and therefore the series diverges.