To determine if the series $\sum_{n=6}^{\infty} \frac{1}{n^2 + 36}$ is convergent, we can apply the Integral Test. The Integral Test states that if $f(x)$ is a positive, continuous, and decreasing function on $[6, \infty)$ and $f(n) = \frac{1}{n^2 + 36}$, then the convergence of $\int_{6}^{\infty} f(x) \, dx$ determines the convergence of the series.

Steps:

1. Define $f(x) = \frac{1}{x^2 + 36}$.
2. Check if $f(x)$ is positive, continuous, and decreasing on $[6, \infty)$. Since $x^2 + 36 > 0$ for all $x$, $f(x)$ meets these conditions.
3. Evaluate the improper integral $\int_{6}^{\infty} \frac{1}{x^2 + 36} \, dx$.

The integral is: $\int \frac{1}{x^2 + 36} \, dx = \frac{1}{6} \arctan\left(\frac{x}{6}\right) + C.$

4. Now, evaluate the improper integral from $6$ to $\infty$: $\int_{6}^{\infty} \frac{1}{x^2 + 36} \, dx = \lim_{b \to \infty} \left[ \frac{1}{6} \arctan\left(\frac{x}{6}\right) \right]_{6}^{b}.$

5. Substitute the limits: $= \lim_{b \to \infty} \left( \frac{1}{6} \arctan\left(\frac{b}{6}\right) - \frac{1}{6} \arctan\left(1\right) \right).$

Since $\arctan\left(\frac{b}{6}\right) \to \frac{\pi}{2}$ as $b \to \infty$, the integral converges to a finite value: $= \frac{1}{6} \cdot \frac{\pi}{2} - \frac{1}{6} \arctan(1) = \frac{\pi}{12} - \frac{1}{6} \cdot \frac{\pi}{4} = \text{finite}.$

Since the integral converges, by the Integral Test, the series $\sum_{n=6}^{\infty} \frac{1}{n^2 + 36}$ also converges.

Would you like further details on any of the steps?

Related Questions

1. What is the Integral Test for convergence, and when is it applicable?
2. How do improper integrals relate to series convergence?
3. Why does $\arctan(x)$ approach $\frac{\pi}{2}$ as $x \to \infty$?
4. Can the Integral Test be used on series that don't start at $n = 1$?
5. What other tests for convergence can be applied to similar series?

Tip

When using the Integral Test, ensure the function is positive, continuous, and decreasing over the interval; otherwise, the test cannot be applied correctly.