The series $\sum_{n=1}^\infty \frac{n}{n+1}$ $$\sum_{n=1}^\infty \frac{n}{n+1}$$
I have $\displaystyle\lim_{n \to{+}\infty}{\frac{n}{n+1}}=\displaystyle\lim_{n \to{+}\infty}{\frac{\frac{n}{n}}{\frac{n}{n}+\frac{1}{n}}}=\displaystyle\lim_{n \to{+}\infty}{\frac{1}{1+\frac{1}{n}}}=1$
Since this is not 0, the series diverges by the divergence test.
Is it right to my result?
 A: 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.
A: Yes. You are right. However, there is no one universal divergence test. You need to say that since $\displaystyle \lim_{n \rightarrow \infty} a_n \neq 0$, we have that $\displaystyle \sum_{n=1}^{\infty} a_n$ doesn't converge.
EDIT As Arturo points out, looks like what I wrote above is "The divergence test". So you are fine provided the question is if the series diverges.
