Absolute convergence of an infinite series and p-series test

Why does the infinite series $\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt n}z^n$ where $z\in \mathbb{C}$ converge absolutely for $|z|<1$. Doesn't the series diverge because if we apply the absolute values, we can use p-series test and since $p<1$, the series diverges? Also, I am trying to find a value of $z$ with $|z|=1$ such that the series converges but I am stuck on this part too.

• For $\lvert z\rvert < 1$, you have the domination by a convergent geometric series. – Daniel Fischer Dec 21 '14 at 21:37

Hint

• Use the ratio test to prove that the radius of convergence is $R=1$.
• Use the Dirichlet's test to prove the convergence for $|z|=1$ and $z\ne1$, and for $z=1$ use the Leibniz criterion to prove the convergence.

No, that's only for ordinary series. You can see by the root test that

$$\lim_{n\to\infty}\left|\sqrt{n}z^n\right|^{1/n}=\lim_{n\to\infty}n^{1/2n}|z|=|z|$$

converges absolutely when this limit is $<1$, i.e. when $|z|<1$.

$$\sum_{n=1}^\infty\left|\frac{(-1)^nz^n}{\sqrt{n}}\right|=\sum_{n=1}^\infty\left|\frac{z^n}{\sqrt{n}}\right|\geq\sum_{n=1}^\infty\left|\frac{z^n}{n}\right|$$ If $$|z|=1$$, the rightmost part becomes the harmonic series: $$\sum_{n=1}^\infty\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+...$$ for which we know diverges. If $$|z|=1$$ diverges, $$|z|>1$$ should diverge as well. It follows that the LHS of the inequality diverges for these values as well.

Now, if $$|z|<1$$, the LHS of the inequality will become a sum of infinite geometric series with common ratio |z|<1 and converges.