Find the value of P such that the seris converge $$\sum_{n=2}^{\infty} \frac{1}{\log^p(n)} \tag{1.}$$
$$\sum_{n=2}^{\infty} e^{n(p+1)} \tag{2.}$$
In 1 if $p=0$ then the whole series is 1.
in 2 I can look at some individual results.
The question is what is the correct way to answer this.

This can be shown by using Ratio Test.
For (1), $$a_n=\ln(n)^{-p}$$
The ratio is, $$\frac{a_{n+1}}{a_n}=(\frac{\ln(n)}{\ln(n+1)})^p$$ 
Thus for any $p\gt 0$, the ratio is less than 1 and the series absolutely converges.
For (2), $$a_n=e^{n(p+1)}$$
The ratio is, $$\frac{a_{n+1}}{a_n}=(\frac{e^{(n+1)}}{e^n})^{(p+1)}=e^{p+1}$$ 
Thus for any $p\lt -1$, the ratio is less than 1 and the series absolutely converges.
