Compute the value of the following improper integral. If it converges, enter its value. Enter infinity if it diverges to $\infty$, and -infinity if it diverges to $-\infty$. Otherwise, enter diverges. \[ \int_{1}^{\infty} 8 x^{2} e^{-x^{3}} d x= \] Does the series $\sum_{n=1}^{\infty} 8 n^{2} e^{-n^{3}}$ converge or diverge? ?

Step 1 ：First, we can rewrite the integral as follows: \[\int_{1}^{\infty} 8 x^{2} e^{-x^{3}} d x = 8 \int_{1}^{\infty} x^{2} e^{-x^{3}} d x\]

Step 2 ：Next, we can use the substitution method to simplify the integral. Let \(u = x^3\), then \(du = 3x^2 dx\). So, \(x^2 dx = \frac{1}{3} du\). The limits of integration also change: when \(x = 1\), \(u = 1\), and when \(x = \infty\), \(u = \infty\).

Step 3 ：So, the integral becomes: \[8 \int_{1}^{\infty} x^{2} e^{-x^{3}} d x = 8 \int_{1}^{\infty} \frac{1}{3} e^{-u} du = \frac{8}{3} \int_{1}^{\infty} e^{-u} du\]

Step 4 ：Now, we can evaluate the integral: \[\frac{8}{3} \int_{1}^{\infty} e^{-u} du = \frac{8}{3} [-e^{-u}]_{1}^{\infty} = \frac{8}{3} [0 - (-e^{-1})] = \frac{8}{3e}\]

Step 5 ：Next, we consider the series \[\sum_{n=1}^{\infty} 8 n^{2} e^{-n^{3}}\]. We can use the comparison test to determine whether it converges or diverges. We compare it with the series \[\sum_{n=1}^{\infty} \frac{8}{n}\], which is a p-series with \(p = 1 < 1\), so it diverges.

Step 6 ：Since \[0 \leq 8 n^{2} e^{-n^{3}} \leq \frac{8}{n}\] for all \(n \geq 1\), and the larger series diverges, by the comparison test, the series \[\sum_{n=1}^{\infty} 8 n^{2} e^{-n^{3}}\] also diverges.

Step 7 ：Finally, we have the value of the integral is \(\boxed{\frac{8}{3e}}\), and the series diverges.