To test the given series

$\sum_{n=1}^\infty \frac{3^n + 5}{6^n},$

we use the Limit Comparison Test and compare it to a simpler geometric series of the form:

$\sum_{n=1}^\infty r^n.$

Key Steps:

1. Dominant term analysis: The dominant term in the numerator is $3^n$, and in the denominator it is $6^n$. Hence, the dominant behavior of the series is determined by the ratio $\frac{3^n}{6^n}$.

2. Simplification: $\frac{3^n}{6^n} = \left(\frac{3}{6}\right)^n = \left(\frac{1}{2}\right)^n.$

   Therefore, the series behaves like the geometric series:

   $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n.$

3. Comparison series: We compare the original series to $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n$, where $r = \frac{1}{2}$.

4. Limit Comparison Test formula: Compute: $\lim_{n \to \infty} \frac{\frac{3^n + 5}{6^n}}{\left(\frac{1}{2}\right)^n}.$

   Simplify: $\frac{\frac{3^n + 5}{6^n}}{\left(\frac{1}{2}\right)^n} = \frac{\frac{3^n + 5}{6^n}}{\frac{3^n}{6^n}} = \frac{3^n + 5}{3^n} = 1 + \frac{5}{3^n}.$

   As $n \to \infty$, $\frac{5}{3^n} \to 0$, so: $\lim_{n \to \infty} \frac{\frac{3^n + 5}{6^n}}{\left(\frac{1}{2}\right)^n} = 1.$

   Since this limit is a finite non-zero constant, the original series converges if and only if the comparison series converges.

5. Conclusion: The series $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n$ is a geometric series with $r = \frac{1}{2} < 1$, which converges. Therefore, the original series also converges.

Final Answer:

The series to compare is: $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n \quad \text{where } r = \frac{1}{2}.$

Would you like me to go into more detail or provide further examples?

Related Questions:

1. How does the Limit Comparison Test work in general?
2. What are other methods to test series convergence?
3. Can you explain the convergence of geometric series in more depth?
4. What happens if $r \geq 1$ in a geometric series?
5. Are there alternative series to compare to in this problem?

Tip:

When applying the Limit Comparison Test, always focus on the dominant terms of the numerator and denominator to find a simpler comparison series.