## Thomas' Calculus 13th Edition

Here, $\Sigma_{n=1}^\infty \dfrac{1}{3^{n-1}+1} \lt \Sigma_{n=1}^\infty\dfrac{1}{3^{n-1}}$ and $\Sigma_{n=1}^\infty\dfrac{1}{3^{n-1}}$ shows a geometric convergent series and has common ratio $r=\dfrac{1}{3} \gt 1$. Hence, the series converges by the direct comparison test.