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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.

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

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$$\sum_{n=1}^\infty\left|\frac{(-1)^nz^n}{\sqrt{n}}\right|=\sum_{n=1}^\infty\left|\frac{z^n}{\sqrt{n}}\right|\geq\sum_{n=1}^\infty\left|\frac{z^n}{n}\right|$$ 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.

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