To determine if the series

$\sum_{n=1}^{\infty} n^7 e^{-n^8}$

is convergent or divergent using the Integral Test, we need to evaluate the corresponding integral:

$\int_1^{\infty} x^7 e^{-x^8} \, dx$

Solution:

1. Set up the integral: $\int_1^{\infty} x^7 e^{-x^8} \, dx$

2. Use substitution to simplify the integral: Let $u = x^8$, so $du = 8x^7 \, dx$, which gives $dx = \frac{du}{8x^7}$.

3. Substitute $x^7 \, dx = \frac{du}{8}$ into the integral: $\int_1^{\infty} x^7 e^{-x^8} \, dx = \frac{1}{8} \int_{1}^{\infty} e^{-u} \, du$

4. Evaluate the integral: $\frac{1}{8} \int_1^{\infty} e^{-u} \, du = \frac{1}{8} \left[ -e^{-u} \right]_1^{\infty}$ $= \frac{1}{8} \left( 0 + e^{-1} \right) = \frac{e^{-1}}{8}$

Since this integral evaluates to a finite value, the integral converges.

Conclusion:

Since the integral converges, the series

$\sum_{n=1}^{\infty} n^7 e^{-n^8}$

is also convergent by the Integral Test.

Would you like more details on any step or have any questions?

Related Questions:

1. What is the Integral Test for series convergence?
2. How does substitution simplify improper integrals?
3. Can the Integral Test determine divergence as well?
4. Why does convergence of the integral imply convergence of the series?
5. What other convergence tests are useful for similar series?

Tip: When using the Integral Test, ensure that the function you integrate is positive, continuous, and decreasing on the interval $[1, \infty)$.