Question

Test each of the following series for convergence.

$\mathbf{}\mathbf{\sum }_{\mathbf{}}{\mathit{e}}^{\mathbf{i}\mathbf{n}\frac{\mathbf{\pi }}{\mathbf{6}}}$

Short answer

The series is divergent.

Step-by-step solution

Given Information

The series is $\underset{}{\sum e^{\frac{in\mathrm{\pi }}{6}}}$.

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.

Test the convergence.

The series is $\underset{}{\sum e^{\frac{in\mathrm{\pi }}{6}}}$.

$$\begin{gathered} A_n=e^{in\mathrm{\pi }/6} \\ {\mathrm{A}}_{\mathrm{n}+1}={\mathrm{e}}^{\mathrm{i}(\mathrm{n}+1)\mathrm{\pi }/6} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{p}}_{\mathrm{n}}=\left|\frac{{\mathrm{A}}_{\mathrm{n}+1}}{{\mathrm{A}}_{\mathrm{n}}}\right| \\ =\frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{n}+1)\mathrm{\pi }/6}}{{\mathrm{e}}^{\mathrm{i}\left(\mathrm{n}\right)\mathrm{\pi }/6}} \\ ={\mathrm{e}}^{\mathrm{i\pi }/6} \end{gathered}$$

Find the limit role="math" localid="1658730767543" $\mathrm{p}=\underset{\mathrm{n}\to \infty }{\lim }\frac{{\mathrm{A}}_{\mathrm{n}+1}}{{\mathrm{A}}_{\mathrm{n}}}$

$$\begin{gathered} \mathrm{p}=\left|{\mathrm{e}}^{\mathrm{i\pi }/6}\right| \\ =\left|\cos \left(\mathrm{\pi }/6\right)+\mathrm{i}\sin \left(\mathrm{\pi }/6\right)\right| \\ =1 \end{gathered}$$

Test fails.

Find the sum of the series.

$$\begin{gathered} \mathrm{S}={\int }_1^{\infty }{\mathrm{e}}^{\mathrm{in\pi }/6}\mathrm{dn} \\ =\frac{6}{\mathrm{i\pi }}{\left[{\mathrm{e}}^{\mathrm{in\pi }/6}\right]}_1^{\infty } \end{gathered}$$

The sum is not defined.

Hence the series is divergent.