## Thomas' Calculus 13th Edition

Here, we have $\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(1)(3) ....(2n-1)(2n+1)}{[(2)(4)....2n(2n+2)](3^{n+1}+1)}}{\dfrac{(1)(3) ....(2n-1)}{[(2)(4)....2n](3^n+1)}}|$ This implies that $\lim\limits_{n \to \infty}|\dfrac{(2n+1)(3^n+1)}{(2n+2)(3^{n+1}+1)}|=\lim\limits_{n \to \infty}|\dfrac{(2+\dfrac{1}{n})(1+\dfrac{1}{3^n})}{(2+\dfrac{2}{n})(3+\dfrac{1}{3^n})}|=\dfrac{(2+0)(1+0)}{(2+0)(3+0)}=\dfrac{1}{3} \lt 1$ Thus, the series converges by the ratio test.