1. **Stating the problem:** We have a sequence defined by the partial sums $$a_n = \frac{1}{3+1} + \frac{1}{3^2+1} + \cdots + \frac{1}{3^n+1}$$ for $n=1,2,\ldots$. We want to determine whether this sequence converges. 2. **Rewrite the terms:** Notice each term in the sequence is of the form $\frac{1}{3^k+1}$ where $k=1,2,\ldots,n$. 3. **Behavior of terms:** As $k$ grows larger, $3^k$ grows exponentially. Therefore, each term $\frac{1}{3^k+1}$ becomes very small quickly. 4. **Test for convergence of the series:** Since the terms are positive and decreasing, and $\frac{1}{3^k+1} < \frac{1}{3^k}$, we compare with the geometric series $\sum_{k=1}^\infty \frac{1}{3^k}$. 5. **Geometric series sum:** The series $\sum_{k=1}^\infty \frac{1}{3^k}$ converges because $|r|= \frac{1}{3} < 1$. Its sum is $$\sum_{k=1}^\infty \frac{1}{3^k} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}.$$ 6. **Comparison test conclusion:** Since $0 < \frac{1}{3^k+1} < \frac{1}{3^k}$ and $\sum \frac{1}{3^k}$ converges, by the comparison test, the series $$\sum_{k=1}^\infty \frac{1}{3^k+1}$$ converges. 7. **Hence, the sequence $a_n$ converges to a finite limit as $n \to \infty$. **Final answer:** The sequence $a_n = \sum_{k=1}^n \frac{1}{3^k+1}$ converges.