Question

Find the disk of convergence for each of the following complex power series.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{z}}^{\mathbf{n}}}{\sqrt{\mathbf{n}}}$

Short answer

The required disk of convergence is .$\left|z\right|<1$

Step-by-step solution

Define Disk of Convergence

For any power series $\sum a_n{z}^n$where z is a complex numbers, then disk of convergence is given by: .$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{\left|}\frac{\mathbf{z}\mathbf{\times }\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{z}\mathbf{\right|}$$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathbf{\left|}\frac{\mathbf{z}\mathbf{\times }\mathbf{n}}{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{z}\mathbf{\right|}$

Step 2:Find the disk of Convergence 

The given power series is: $\sum _{n=1}^{\infty }\frac{z^n}{\sqrt{n}}$, where$a_n=\frac{z^n}{\sqrt{n}}$

Now, let us evaluate the ratio as:

$\begin{array}{c}\rho =\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{\frac{z^{n+1}}{\sqrt{n+1}}}{\frac{z^n}{\sqrt{n}}}\right|\\ =\underset{n\to \infty }{\lim }\left|z\sqrt{\frac{n}{n+1}}\right|\end{array}$

Now, for the series to be convergent, we have$\rho <1$ . So,

$\begin{array}{l}\rho =\underset{n\to \infty }{\lim }\left|z\sqrt{\frac{n}{n+1}}\right|<1\\ \left|z\right|\underset{n\to \infty }{\lim }\left|\sqrt{\frac{n}{n+1}}\right|<1\\ \left|z\right|<1\end{array}$

Hence, the required disk of convergence is .$\left|z\right|<1$