To determine whether the series

$\sum_{n=1}^{\infty} n^3 e^{-n^4}$

is convergent or divergent using the Integral Test, we evaluate the improper integral

$\int_{1}^{\infty} x^3 e^{-x^4} \, dx.$

Step 1: Set up the Integral

The function we need to integrate is $f(x) = x^3 e^{-x^4}$.

Step 2: Use a Substitution

Let $u = x^4$. Then, $\frac{du}{dx} = 4x^3$, or equivalently $dx = \frac{du}{4x^3}$.

When $x = 1$, $u = 1^4 = 1$. When $x \to \infty$, $u \to \infty$.

Substituting into the integral, we get:

$\int_{1}^{\infty} x^3 e^{-x^4} \, dx = \int_{1}^{\infty} e^{-u} \cdot \frac{du}{4} = \frac{1}{4} \int_{1}^{\infty} e^{-u} \, du.$

Step 3: Evaluate the Integral

Now we have:

$\frac{1}{4} \int_{1}^{\infty} e^{-u} \, du.$

The integral of $e^{-u}$ is $-e^{-u}$, so we get:

$\frac{1}{4} \left[ -e^{-u} \right]_{1}^{\infty} = \frac{1}{4} \left( 0 + e^{-1} \right) = \frac{1}{4e}.$

Conclusion

Since the integral converges to a finite value, by the Integral Test, the series $\sum_{n=1}^{\infty} n^3 e^{-n^4}$ is convergent.

Would you like further details on each step or have any questions?

Follow-Up Questions:

1. How does the Integral Test work, and why does it imply convergence?
2. What other convergence tests could we use for this series?
3. Could a comparison test be applied to this problem?
4. Why is the substitution $u = x^4$ helpful in evaluating this integral?
5. How does this series compare to a simpler exponential series?

Tip:

When using the Integral Test, check that the function $f(x)$ is positive, continuous, and decreasing for $x \geq 1$ to apply the test correctly.