Question

Find Sum of power series ,then answer the questions
1- series from 1 to infinity of [ {(-1)^(n+1)} .{ n} . { (z-1)^n} ] indicate the convergence nhd (neighborhood).

is series from 1 to infinity of [ n. {(1-4i)^n} convergent if yes compute its sum

Answer 1

Answer on Question #49961 – Math – Complex Analysis

1) Given:

$$\sum_{n=1}^{\infty} (-1)^{n+1} n (z - 1)^n$$

Solution:

$c_n = (-1)^{n+1} n$ is the nth term

$R$ is a radius of convergence

$U_R$ is the convergence neighborhood

$$R = \frac{1}{\lim_{n \to \infty} \sqrt[n]{|c_n|}} = \frac{1}{\lim_{n \to \infty} \sqrt[n]{|n|}} = 1$$

then $U_R = \{z : |z - 1| < 1\}$

Answer: $R = 1$

$$U_R = \{z : |z - 1| < 1\}$$

2) Given:

$$\sum_{n=1}^{\infty} n (1 - 4i)^n$$

Solution:

Assume that the initial series is convergent.

The series $\sum_{n=1}^{\infty} n(1 - 4i)^n = (1 - 4i) \sum_{n=1}^{\infty} n(1 - 4i)^{n-1}$ is the derivative of $(1 - 4i) \sum_{n=1}^{\infty} (1 - 4i)^n$.

Convergent series are differentiable. The last series is convergent only if $|1 - 4i| < 1$, which is false, because $|1 - 4i| = \sqrt{17} > \sqrt{16} = 4$, i.e. $|1 - 4i| > 1$. We have obtained a contradiction, our assumption was not correct. Thus, the series $\sum_{n=1}^{\infty} n(1 - 4i)^n$ diverges.

Answer: not convergent

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