University Calculus: Early Transcendentals (3rd Edition)

Consider $a_n=\dfrac{n+2^n}{n^2 2^{n}}$ and $b_n=\dfrac{1}{ n^2}$ Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{n+2^n}{n^2 2^{n}}}{\dfrac{1}{ n^2}}$ Thus, we have $=\lim\limits_{n \to \infty} \dfrac{n+2^n}{2^n}=\lim\limits_{n \to \infty} 1+\frac{n}{2^n}$ Apply L-Hospital's rule, we get $=1+\lim\limits_{n \to \infty} \dfrac{1}{2^n\ln 2}=1+0=1$ Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{2}}$ is a convergent series due to the p-series test with $p=2 \gt 1$ Thus the series converges by the limit comparison test.