For Series of Real Numbers, we have the following theorem. If $\{a_n\}$ and $\{b**_n**\}$ be two sequences of positive real numbers such that $0\geq a_n \geq k\cdot b_n$, for some real number $k$ then
- $\{b_n\}$ converges $\implies$ $\{a_n\}$ converges.
- $\{a_n\}$ diverges $\implies$ $\{b_n\}$ diverges.
For Series of complex Numbers, we have no order, hence we have to modify our statement accordingly.
Let $\{z_n\}$ and $\{w_n\}$ be two sequences of complex numbers.
Let, $|z_n|\leq k |w_n|$ then
- $\sum |w_n|$ is convergent implies $\sum z_n$ is absolutely convergent.
- *My Question is: "Can we say that $\sum |z_n|$ is divergent implies $\sum w_n$ is divergent."? * (What I know is $\sum |z_n|$ is divergent implies $\sum |w_n|$ is not absolutely convergent.)
Further Question:
I have a further question: If the statement above is not true then how do we show that " If a power series $\sum a_nz^n$ is not divergent at $z=z_0$, then it is divergent for all $z$ satisfying $|z|>|z_0|$"
Best Answer
No, we cannot. Take $z_n=\dfrac1n$, $w_n=\dfrac{(-1)^n}n$, and $k=1$.