To prove the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{2^n + 3^n}$ using the comparison test, we will compare it to a simpler series.

1. Identify a simpler series: Note that for $n \geq 1$, $3^n$ grows faster than $2^n$. Therefore, we can say that:

   $2^n + 3^n \geq 3^n$

   This leads us to:

   $\frac{1}{2^n + 3^n} \leq \frac{1}{3^n}$

2. Consider the series $\sum \frac{1}{3^n}$: This is a geometric series with a common ratio $r = \frac{1}{3}$. A geometric series converges if $|r| < 1$. In this case, since $|\frac{1}{3}| < 1$, the series $\sum_{n=1}^{\infty} \frac{1}{3^n}$ converges.

3. Apply the comparison test: Since $\frac{1}{2^n + 3^n} \leq \frac{1}{3^n}$ for all $n \geq 1$ and $\sum \frac{1}{3^n}$ converges, we can apply the comparison test. Thus, we conclude:

   $\sum_{n=1}^{\infty} \frac{1}{2^n + 3^n} \text{ converges.}$

Conclusion

By the comparison test, the series $\sum_{n=1}^{\infty} \frac{1}{2^n + 3^n}$ converges.

Would you like further details on any specific part of the proof, or do you have any questions? Here are some follow-up questions you might consider:

1. What are the conditions for the comparison test to be valid?
2. Can you explain how to determine the convergence of geometric series?
3. What other convergence tests can be applied to series?
4. How does the behavior of the terms affect convergence?
5. Could you demonstrate the divergence of a similar series?

Tip: Always check if the series you are comparing to is known to converge or diverge before applying the comparison test.