Convergence test for series Let  $\space \displaystyle\sum_{n=1}^{\infty}  \space \left | \frac{\cos(n\pi)}{n+1}\right |$. Does this series converge or not?


The serie is valid for the natural numbers, so it can be writen as $\space \displaystyle\sum_{n=1}^{\infty}  \space  \frac{\left |\cos(n\pi)\right |}{n+1}$.

One knows that $\space 0\leq\left |\cos(n\pi)\right |\leq 1$. By using the asymptotic concept, one can say that the leader terms of the top and bottom of the fraction are $\displaystyle \frac{1}{n}$.

And so, the original serie is equivalent to this one
$$\space \displaystyle\sum_{n=1}^{\infty}  \space \frac{1}{n}$$

that is a $p$ series, where $p\leq 1$ and so diverges, and also the original series. This is correct? Thanks.
No. Since the p-series you've considered was only an upper bound for the series, the fact that it diverges to infinity tells you nothing.
Since $|\cos(n\pi)| = 1\;\; \forall n,\;$ your series is equivalent to:

$$\sum_{n = 1}^{\infty}\frac{\left|(-1)^n\right|}{(n + 1)} = \sum_{n = 1}^{\infty} \frac {1}{n+1} $$

You're "spot on" with respect to comparing your given series to the harmonic series: $\;\;\displaystyle \sum_{n = 1}^\infty \dfrac 1n,\;$ but since $\;\dfrac 1{n+1} \lt \dfrac 1n,\;$ you can note that 

$$\sum_{\color{blue}{\bf n = 1}}^\infty \frac 1{n+1} \quad = \quad \sum_{\color{blue}{\bf n = 2}}\frac 1n$$ and so, save for one term, your series is the harmonic series, and diverges just as does the "strict" harmonic series. 

If you want a tighter argument, you can use the limit comparison test of your series and the "strict" harmonic series to prove your series diverges.

$$\text{Put, say}\;\;a_n = \dfrac 1{n+1},\;\;b_n = \dfrac 1n \iff
\dfrac{a_n}{b_n} = \dfrac{1/(n+1)}{1/n} = \dfrac n{n+1}$$

$$\text{Then we have}\quad\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac n{n+1} = 1 &lt; \infty$$

Since the harmonic series given by $\sum b_n$ diverges, so too does your series.