I have $\sum(\frac{(-1)^n}{\sqrt{n}} + \frac1n)$.
Nth test: $\lim_{n->\infty}{(\frac{(-1)^n}{\sqrt{n}} + \frac1n)}$ = 0.
I think that we can not split it into two sums like $\sum(\frac{(-1)^n}{\sqrt{n}}) + \sum(\frac1n)$ because the second one is divergence.
I know that sum is absolutely divergence but how can I prove relative divergence? All tests say about the sum convergence but not divergence.
The pertinent theorem is that if $\sum a_n$ and $\sum b_n$ are both convergent, then so is $\sum(a_n+b_n)$. This theorem doesn't care whether the convergence is conditional or absolute.
So if $\sum\left({(-1)^n\over\sqrt n}+{1\over n}\right)$ were convergent, then, since $-\sum{(-1)^n\over\sqrt n}$ is convergent, we would have to conclude that $\sum\left({(-1)^n\over\sqrt n}+{1\over n}-{(-1)^n\over\sqrt n}\right)=\sum{1\over n}$ is convergent, which we know not to be the case.
The key to proving the pertinent theorem is to note that if $\left|\sum_{n=1}^Na_n-A\right|\lt\epsilon/2$ and $\left|\sum_{n=1}^Nb_n-B\right|\lt\epsilon/2$, then
$$\left|\sum_{n=1}^N(a_n+b_n)-(A+B)\right|\le\left|\sum_{n=1}^Na_n-A\right|+\left|\sum_{n=1}^Nb_n-B\right|\lt\epsilon/2+\epsilon/2=\epsilon$$