Absolute convergence of an infinite series and p-series test

686 Views Asked by At

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.

2

There are 2 best solutions below

0
On BEST ANSWER

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.
0
On

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 $|z|<1$.