$(Given)\ n^{th}\ term =u_n= \dfrac {2^n+ 1} {(3^n +n)}\ u_{n+1}= \dfrac {2^{n+1}+ 1} {3^{n+1} +(n+1)} Hence,\ $ $\dfrac {u_n} {u_{n+1}}= \dfrac {2^n+ 1} {(3^n +n)} \times \dfrac {3^{n+1}+ (n+1)} {2^{n+1} +1}\ = \dfrac {2^n+ 1} {(2^n.2^n +1)} \times \dfrac {{3^n.3}+ (n+1)} {3^ n+n}$ $\lim_{n \to \infty} \dfrac {u_{n+1}} {u_n}= \lim_{n \to \infty} \bigg[\dfrac {2^n(1+ 1/2^n)} {(2^n (2+1/2^n)} \times \dfrac {3^n [3+(n+1)/3^n]} {3^n [1+n/3^n]} \bigg]$ $= \lim_{n \to \infty} \dfrac {(1+ 1/2^n)} { (2+1/2^n)} \times \lim_{n \to \infty}\dfrac {3+n/3^n+1/3^n} {1+n/3^n}$ $\lim_{n \to \infty} \dfrac 1 {3^n} = \lim_{n \to \infty}\dfrac {1} {2^n}=0\ and\ \lim_{n \to \infty} \dfrac n {3^n} = \lim_{n \to \infty}\dfrac {1} {3^n.log3}=0\ [Using\ L'Hospital's\ rule]\ $

(\lim_{n \to \infty}\dfrac {u_{n+1}} {u_{n}}= \dfrac {1+ 0} {(2 +0)} \times \dfrac {3+ 0+0} {1 +0}\ =\dfrac {3} 2 > 1.\ [By\ D'Alembert's\ ratio\ test,]\ \sum u_n= {\sum (2^n+1})/{(3^n +n)}\ is\ convergent.\)