Thomas' Calculus 13th Edition

Let $u_n=\dfrac{n+2^n}{n^2 2^{n}}$ and $v_n=\dfrac{1}{ n^2}$ Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{n+2^n}{n^2 2^{n}}}{1/ n^2}$ $\implies \lim\limits_{n \to \infty} \dfrac{n+2^n}{2^n}=\dfrac{\infty}{\infty}$ Need to apply L-Hospital's rule. $1+\lim\limits_{n \to \infty} (\dfrac{1}{2^n\ln 2})=1+0=1$ But $\Sigma_{n=1}^\infty \dfrac{1}{n^{2}}$ shows a convergent $p$-series with $p=2 \gt 1$ Hence, the series converges by the limit comparison test.