Q7P - Mathematical Methods in Physical Sciences

Question

Test each of the following series for convergence.

$\underset{\mathbf{ }}{\mathbf{\sum }\frac{{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{i}\mathbf{)}}^{\mathbf{n}}}{\mathbf{n}}}$

Step-by-Step Solution

Answer

The series is divergent.

Step 1: Given Information

The series is $\underset{ }{\sum \frac{{(1-\mathrm{i})}^{\mathrm{n}}}{\mathrm{n}}}$.

Step 2: Definition of the Convergent series.

A series will be considered to be convergent in the condition that if terms in the series move towards infinity then the terms move to the zero value.

Step 3: Test the convergence.

The series is $\underset{ }{\sum \frac{{(1-\mathrm{i})}^{\mathrm{n}}}{\mathrm{n}}}$.

 $$\begin{gathered} {\mathrm{A}}_{\mathrm{n}}=\underset{ }{\frac{{(1-\mathrm{i})}^{\mathrm{n}}}{\mathrm{n}}} \\ {\mathrm{A}}_{\mathrm{n}+1}=\underset{ }{\frac{{(1-\mathrm{i})}^{\mathrm{n}+1}}{\mathrm{n}+1}} \end{gathered}$$

 

Find the ratio $\left|\frac{{\mathrm{A}}_{\mathrm{n}+1}}{{\mathrm{A}}_{\mathrm{n}}}\right|$.

$$\begin{gathered} {\mathrm{p}}_{\mathrm{n}}=\frac{{\mathrm{A}}_{\mathrm{n}+1}}{{\mathrm{A}}_{\mathrm{n}}} \\ =\frac{{(\mathrm{i}-1)}^{\mathrm{n}+1}}{\mathrm{n}+1}\times \frac{\mathrm{n}}{{(\mathrm{i}-1)}^{\mathrm{n}}} \\ =\frac{(\mathrm{i}-1)}{1+{\displaystyle \frac{1}{\mathrm{n}}}} \end{gathered}$$ 

 

Find the limit $\mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\frac{{\mathrm{A}}_{\mathrm{n}+1}}{{\mathrm{A}}_{\mathrm{n}}}$ 

$$\begin{gathered} \mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\frac{{\mathrm{A}}_{\mathrm{n}+1}}{{\mathrm{A}}_{\mathrm{n}}} \\ \mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{(\mathrm{i}-1)}{1+{\displaystyle \frac{1}{\mathrm{n}}}}\right| \\ =\left|(\mathrm{i}-1)\right| \\ =\sqrt{2} \end{gathered}$$ 

 

 p > 1, Hence the series is divergent.