# Convergence domain of $\sum_\limits{n=1}^{\infty}(-1)^{n+1}\frac{1}{n^x}$?

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!

• The Dirichlet eta function (which is essentially the same function presented here), converges absolutely on $Re(x) > 1$. It turns out it still converges on $0 \lt x \le 1$, but only conditionally, meaning that in that small strip its counterpart $\sum_{n=1}^\infty \frac{1}{n^x}$ (the Riemann zeta function) diverges. Under conditional convergence, an expression, after rearranging of terms, can converge to an arbitrarily large number or even diverge. – Bladewood Apr 13 '18 at 10:32

## 1 Answer

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 $\lim_{n\to \infty} |a_n|\neq 0$.