Here is another way of looking at it:
$\sum_{n=1} ^{\infty} \frac{1}{n+1} \leq \sum_{n=1} ^{\infty} \frac{n}{n+1}$
since $\frac{1}{n+1} \leq \frac{n}{n+1}$ for all $n>0$. Clearly $\sum_{n=1} ^{\infty} \frac{1}{n+1}$ diverges since it is the harmonic series minus $1$, and therefore $\sum_{n=1} ^{\infty} \frac{n}{n+1}$ diverges.
As t.b. points out in the comments there is a more basic way to bound this series. Notice that
$\frac{1}{2}\leq \frac{n}{n+1}$
for all positive $n$. Therefore
$\sum_{n=1} ^{\infty} \frac{1}{2} \leq \sum_{n=1} ^{\infty} \frac{n}{n+1}.$
Since $\sum_{n=1} ^{\infty} \frac{1}{2}$ is divergent, $\sum_{n=1} ^{\infty} \frac{n}{n+1}$ must diverge.