Q10P - Mathematical Methods in Physical Sciences

Question

Test each of the following series for convergence.

$\underset{\mathbf{ }}{\mathbf{\sum }{\left(\frac{1+i}{1-i\sqrt{3}}\right)}^{\mathbf{n}}}$

Step-by-Step Solution

Answer

The series is convergent.

Step 1: Given Information

The series is $\sum _{ }{\left(\frac{1+\mathrm{i}}{1-\mathrm{i}\sqrt{3}}\right)}^{\mathrm{n}}$.

Step 2: Definition of the Convergent series.

A series is said to be convergent if the terms of a series get close to zero when the number of terms moves towards infinity.

Step 3: Test the convergence.

The series is $\sum _{ }{\left(\frac{1+\mathrm{i}}{1-\mathrm{i}\sqrt{3}}\right)}^{\mathrm{n}}.$.

$$\begin{gathered} {\mathrm{A}}_{\mathrm{n}}=\sum _{ }{\left(\frac{1+\mathrm{i}}{1-\mathrm{i}\sqrt{3}}\right)}^{\mathrm{n}} \\ {\mathrm{A}}_{\mathrm{n}+1}=\sum _{ }{\left(\frac{1+\mathrm{i}}{1-\mathrm{i}\sqrt{3}}\right)}^{\mathrm{n}+1} \end{gathered}$$ 

 

Find the limit $\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{{\mathrm{A}}_{\mathrm{n}+1}}{{\mathrm{A}}_{\mathrm{n}}}\right|$.

$$\begin{gathered} \mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{{\mathrm{A}}_{\mathrm{n}+1}}{{\mathrm{A}}_{\mathrm{n}}}\right| \\ =\underset{\mathrm{n}\to \infty }{\lim }\left|{\left(\frac{1+\mathrm{i}}{1-\mathrm{i}\sqrt{3}}\right)}^{\mathrm{n}+1}\times {\left(\frac{1-\mathrm{i}\sqrt{3}}{1+\mathrm{i}}\right)}^{\mathrm{n}}\right| \\ =\underset{\mathrm{n}\to \infty }{\lim }\left|\left(\frac{1+\mathrm{i}}{1-\mathrm{i}\sqrt{3}}\right)\right| \end{gathered}$$ 

 

 p < 1, Hence the series is convergent.