Answer
Converges
Work Step by Step
Here, we have the given series as $\Sigma_{n=1}^{\infty} \dfrac{1}{3^n+1}$
Let us consider that $a_n=\Sigma_{n=1}^{\infty} \dfrac{1}{3^n+1}$ and $b_n=\Sigma_{n=1}^{\infty} \dfrac{1}{3^n}$
We can see that the series $b_n$ shows convergent geometric series with $r=\dfrac{1}{3}$.
Next, we have $\lim\limits_{n \to \infty} \dfrac{a_n}{b_n} =\lim\limits_{n \to \infty} \dfrac{\dfrac{1}{3^n+1}}{1/3^n}$
or, $=\lim\limits_{n \to \infty} \dfrac{3^n}{3^n+1}$
or, $=\lim\limits_{n \to \infty} \dfrac{1}{1+1/3^n}$
or, $=1$
Hence, we can see that the given series converges by the limit comparison test.
