## Calculus: Early Transcendentals 8th Edition

The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Given: $\Sigma _{n=3}^{\infty}\frac{{n+2}}{(n+1)^{3}}$ It can be re-written as: $\Sigma _{n=3}^{\infty}\frac{{n+2}}{(n+1)^{3}}=\Sigma _{n=3}^{\infty}\frac{{n+1+1}}{(n+1)^{3}}$ $=\Sigma _{n=3}^{\infty}\frac{{n+1}}{(n+1)^{3}}+\Sigma _{n=3}^{\infty}\frac{{1}}{(n+1)^{3}}$ Adding and subtracting a finite number of terms from a series does not affect the convergence or divergence of the series. The first series is convergent because a $p-$series with $p=2$ is convergent. The second series is convergent because a $p-$series with $p=3$ is convergent. The sum of two convergent series is convergent.