To determine the convergence of the given series

$\sum_{n=7}^{\infty} \frac{n^2}{e^{n/8}}$

we can apply the integral test. This test states that if $f(n)$ is a positive, continuous, and decreasing function for $n \geq N$, then the convergence of the series $\sum_{n=N}^{\infty} f(n)$ is equivalent to the convergence of the improper integral $\int_{N}^{\infty} f(x) \, dx$.

In this case, let:

$f(n) = \frac{n^2}{e^{n/8}}$

Step 1: Set up the Improper Integral

We need to evaluate the integral:

$\int_{7}^{\infty} \frac{x^2}{e^{x/8}} \, dx$

Step 2: Evaluate the Integral

To determine the convergence of this integral, we can perform a substitution. Let $u = \frac{x}{8}$, then $x = 8u$, and $dx = 8 \, du$. Substituting these values into the integral:

$\int \frac{x^2}{e^{x/8}} \, dx = \int \frac{(8u)^2}{e^u} \cdot 8 \, du = 512 \int \frac{u^2}{e^u} \, du$

Now, we need to analyze the behavior of $\int \frac{u^2}{e^u} \, du$ as $u \to \infty$.

The integral $\int \frac{u^2}{e^u} \, du$ converges because the exponential function $e^u$ in the denominator grows faster than the polynomial $u^2$ in the numerator, causing the integral to converge.

Conclusion

Since the improper integral converges, by the integral test, the series

$\sum_{n=7}^{\infty} \frac{n^2}{e^{n/8}}$

is also convergent.

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

Here are five related questions that expand on this topic:

1. How does the comparison test relate to the integral test for series convergence?
2. What are other common convergence tests used for series, and when are they applicable?
3. Can you provide an example of a divergent series and explain why it diverges?
4. How does the rate of growth of the exponential function affect the convergence of series?
5. What would happen if we changed the exponent in the denominator to another power?

Tip: When working with series involving exponentials, checking the relative growth rates of polynomial terms and exponential terms can often indicate convergence or divergence.