How do you use the direct Comparison test on the infinite series #sum_(n=2)^oon^3/(n^4-1)# ?

To use the Direct Comparison Test on the infinite series $\sum_{n=2}^\infty \frac{n^3}{n^4 - 1}$, we need to find a series with terms that are easier to evaluate and that are always greater than or equal to the terms of the given series.

Notice that for all $n \geq 2$,
$n^4 - 1 > n^4 - n^4 = 0$
Thus,
$\frac{n^3}{n^4 - 1} < \frac{n^3}{0}$
is undefined. However, let's look at a slightly modified series: $\sum_{n=1}^\infty \frac{n^3}{n^4}$.

For this series, we have:
$\frac{n^3}{n^4} = \frac{1}{n}$

The series $\sum_{n=1}^\infty \frac{1}{n}$ is a known divergent series, the harmonic series.

Now, since $\frac{n^3}{n^4 - 1}$ is less than $\frac{n^3}{n^4}$ for all $n \geq 2$, and $\sum_{n=1}^\infty \frac{1}{n}$ diverges, we can conclude by the Direct Comparison Test that $\sum_{n=2}^\infty \frac{n^3}{n^4 - 1}$ also diverges.

To use the Direct Comparison Test on the infinite series $\sum_{n=2}^\infty \frac{n^3}{n^4 - 1}$, we need to compare it with a known series whose convergence behavior is understood.

First, we observe that for large values of $n$, $n^4 - 1$ is approximately $n^4$. Thus, we can compare the given series with the series $\sum_{n=2}^\infty \frac{n^3}{n^4} = \sum_{n=2}^\infty \frac{1}{n}$.

The series $\sum_{n=2}^\infty \frac{1}{n}$ is a p-series with $p = 1$. It is a known result that this series diverges (Harmonic Series).

Since $\frac{n^3}{n^4 - 1} < \frac{1}{n}$ for all $n \geq 2$, and the series $\sum_{n=2}^\infty \frac{1}{n}$ diverges, by the Direct Comparison Test, the series $\sum_{n=2}^\infty \frac{n^3}{n^4 - 1}$ also diverges.