## Multivariable Calculus, 7th Edition

If $\lim\limits_{n \to \infty}a_{n}$ does not exist or if $\lim\limits_{n \to \infty}a_{n} \ne 0$, then the series $\Sigma^{\infty}_{n=1}a_{n}$ is divergent. $\Sigma^{\infty}_{k=1} \frac{k(k+2)}{(k+3)^{3}}$ $a_{k}=\frac{k(k+2)}{(k+3)^{3}}$ $\lim\limits_{k \to \infty}a_{k} = \lim\limits_{k \to \infty}\frac{k(k+2)}{(k+3)^{3}}$ $=\lim\limits_{k \to \infty} \frac{1+\frac{2}{k}}{(1+\frac{3}{k})^{2}}$ $=\frac{1+0}{(1+0)^{2}}$ since $\lim\limits_{k \to \infty}\frac{1}{k}=0$ $=1$ $\ne 0$ Since $\lim\limits_{n \to \infty}a_{n} \ne 0$ by the test for divergence, the series is divergent.