Exercise: Determine the convergence domain if the following series $$\sum_\limits{n=1}^{\infty}(-1)^{n+1}\frac{1}{n^x}$$.
Solution: If $x>1$ the series converge absolutely, if $0<x\leqslant 1$ the series converge but not absolutely, if $x\leqslant 0$ the series diverge.
OP attempted resolution:
I tried to solve it in the following way: By the comparison(Weierstrass test) test:$\sum_\limits{n=1}^{\infty}|(-1)^{n+1}\frac{1}{n^x}|\leqslant\sum_\limits{n=1}^{\infty}|1\frac{1}{n^x}|=\sum_\limits{n=1}^{\infty}|\frac{1}{n^x}|$.
Now applying the integral test:
$\int_\limits{1}^{\infty}|\frac{1}{n^x}|=\int_\limits{1}^{\infty} \frac{1}{n^x}<\infty$ if $x<1$, which explains the first part of the solution. However I cannot understand the second part of the solution "$0<x\leqslant 1$ the series converge but not absolutely" either why the series diverge when $x<0$, thought they diverged if $x\leqslant 1$.
Questions:
1) What am I doing wrong? How do I prove "$0<x\leqslant 1$ the series converge but not absolutely"? What does it mean?
2) According to solution "$x\leqslant 0$ the series diverge". Is not when $x\leqslant 1$?
Thanks in advance!
For $0< x \le 1$ the series converges by Alternating series test and diverges absolutely by comparison test with $\sum \frac1 n$.
For $x\le 0$ it diverges since $\lim_{n\to \infty} |a_n|\neq 0$.