To determine the convergence of the series

$\sum_{n=3}^{\infty} 8n e^{-n^2}$

using the Integral Test, let’s go through the following steps:

Step 1: Set up the corresponding function

For the Integral Test, we need a continuous, positive, and decreasing function $f(x)$ that resembles our series' general term. Here, let

$f(x) = 8x e^{-x^2}.$

This function is continuous, positive, and decreasing for $x \geq 3$, so we can proceed with the test.

Step 2: Apply the Integral Test

The Integral Test states that if the improper integral

$\int_{3}^{\infty} 8x e^{-x^2} \, dx$

converges, then the series converges as well. If the integral diverges, the series also diverges.

Step 3: Evaluate the Improper Integral

Let’s evaluate the integral:

$\int_{3}^{\infty} 8x e^{-x^2} \, dx.$

We can solve this by using a substitution. Set $u = -x^2$, then $du = -2x \, dx$, which implies $x \, dx = -\frac{1}{2} \, du$.

Thus,

$\int 8x e^{-x^2} \, dx = \int 8 \cdot -\frac{1}{2} e^{u} \, du = -4 \int e^{u} \, du = -4 e^{u}.$

Substituting back $u = -x^2$, we get

$-4 e^{-x^2}.$

Step 4: Evaluate the Limits

Now we evaluate the improper integral:

$\int_{3}^{\infty} 8x e^{-x^2} \, dx = \left[ -4 e^{-x^2} \right]_{3}^{\infty}.$

As $x \to \infty$, $e^{-x^2} \to 0$, so the upper limit becomes 0. Evaluating at $x = 3$:

$-4 e^{-3^2} = -4 e^{-9}.$

Thus,

$\int_{3}^{\infty} 8x e^{-x^2} \, dx = 0 + 4 e^{-9} = 4 e^{-9}.$

Since this is a finite value, the integral converges.

Conclusion

Since the integral converges, by the Integral Test, the series

$\sum_{n=3}^{\infty} 8n e^{-n^2}$

also converges.

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

Here are five related questions to deepen understanding:

1. How would the approach change if the series started at $n = 1$ instead of $n = 3$?
2. What if the series was $\sum_{n=3}^{\infty} n e^{-n}$ instead? How would you test convergence?
3. Can the Integral Test be applied to any series, or are there specific conditions?
4. What other convergence tests might work for this series, if any?
5. Why is $e^{-n^2}$ more effective at making the series converge than $e^{-n}$?

Tip: The Integral Test works best for series that resemble integrable functions that decrease steadily as $n$ increases.