Question

Test each of the following series for convergence.

$\mathbf{\sum }_{\mathbf{}}\frac{{\mathbf{i}}^{\mathbf{n}}}{\mathbf{n}}$

Short answer

The series is convergent.

Step-by-step solution

Given Information

The series is $\sum _{}\frac{{\mathrm{i}}^{\mathrm{n}}}{\mathrm{n}}$.

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 $\sum _{}\frac{{\mathrm{i}}^{\mathrm{n}}}{\mathrm{n}}$.

For ${\left(\mathrm{i}\right)}^{\mathrm{n}}=1$.

Where n = 4,8,12,16,...4K..

${\mathrm{S}}_1=\sum _{}\frac{1}{4\mathrm{k}}$

For ${\left(\mathrm{i}\right)}^{\mathrm{n}}=-1$.

Where n = 6,10,14,...,4K+2.

${\mathrm{S}}_2=-\sum _{}\frac{1}{4\mathrm{k}+2}$

For ${\left(\mathrm{i}\right)}^{\mathrm{n}}=\mathrm{i}$.

Where n = 5,9,13...,4k+1.

${\mathrm{S}}_3=\mathrm{i}\sum _{}\frac{1}{4\mathrm{k}+1}$

For ${\left(\mathrm{i}\right)}^{\mathrm{n}}=-\mathrm{i}$.

Where n = 7,11,15,...4k+3.

${\mathrm{S}}_4=-\underset{}{\mathrm{i}\sum }\frac{1}{4\mathrm{k}+3}$

The value of the series becomes as follows.

$$\begin{gathered} \mathrm{S}={\mathrm{S}}_1+{\mathrm{S}}_2+{\mathrm{S}}_3+{\mathrm{S}}_4 \\ =\underset{}{\sum \frac{1}{4\mathrm{k}}-\sum \frac{1}{4\mathrm{k}+2}+\mathrm{i}\sum \frac{1}{4\mathrm{k}+1}-\mathrm{i}\sum \frac{1}{4\mathrm{k}+3}} \\ =\underset{}{\sum \frac{1}{4\mathrm{k}}-\frac{1}{4\mathrm{k}+2}+\mathrm{i}\sum \frac{1}{4\mathrm{k}+1}-\frac{1}{4\mathrm{k}+3}} \\ =\sum \frac{1}{4\mathrm{k}(2\mathrm{k}+1)}+\mathrm{i}\sum \frac{1}{(2\mathrm{k}+3)(2\mathrm{k}+1)} \end{gathered}$$

The real part is $\underset{}{\sum \frac{1}{2k(2k+1)}}$.

$$\begin{gathered} \mathrm{R}={\int }_1^{\infty }{\mathrm{a}}_{\mathrm{k}}\mathrm{dk} \\ ={\int }_1^{\infty }\frac{1}{2\mathrm{k}(2\mathrm{k}+1)}\mathrm{dkn} \\ =\frac{1}{2}{\int }_1^{\infty }\frac{1}{{\mathrm{k}}^2(2+1/\mathrm{k})}\mathrm{dk} \end{gathered}$$

Substitute the values given below.

$$\begin{gathered} \mathrm{u}=2+\frac{1}{\mathrm{k}} \\ \mathrm{du}=-{\mathrm{k}}^{-2}\mathrm{dk} \end{gathered}$$

Lower and upper li it becomes as follows.

$$\begin{gathered} {\mathrm{U}}_{\mathrm{u}}=2+\frac{1}{\infty } \\ =2 \\ {\mathrm{U}}_{\mathrm{l}}=2+\frac{1}{1} \\ =3 \end{gathered}$$

The integral becomes as follows.

$$\begin{gathered} \mathrm{R}=-\frac{1}{2}{\int }_3^2\frac{\mathrm{du}}{\mathrm{u}} \\ =-\frac{{\left[\mathrm{Lnu}\right]}_3^2}{2} \\ =\frac{\ln \left(3\right)-\mathrm{In}\left(2\right)}{2} \\ =0.2 \end{gathered}$$

The imaginary part is $$\begin{gathered} \underset{}{\sum ={\int }_1^{\infty }}{\mathrm{b}}_{\mathrm{k}}\mathrm{dk} \\ =2{\int }_1^{\infty }\frac{1}{(4\mathrm{k}+1)(4\mathrm{k}+3)}\mathrm{dk} \end{gathered}$$.

Substitute the values given below.

u = 4k + 1

du = 4 dk

Lower and upper li it becomes as follows.

$$\begin{gathered} {\mathrm{U}}_{\mathrm{u}}=\infty \\ {\mathrm{U}}_{\mathrm{l}}=4+1 \\ =5 \end{gathered}$$

The integral becomes as follows.

$$\begin{gathered} \mathrm{l}=2{\int }_5^{\infty }\frac{\mathrm{du}}{4\mathrm{u}(\mathrm{u}+2)} \\ =\frac{1}{2}{\int }_5^{\infty }\frac{\mathrm{du}}{{\mathrm{u}}^2(1+2/\mathrm{k})} \\ \mathrm{l}=0.08 \end{gathered}$$

The value of a complex number becomes as follows.

$$\begin{gathered} \mathrm{C}=\left|\mathrm{R}+\mathrm{il}\right| \\ =\left|0.2+\mathrm{i}0.08\right| \\ =0.215 \end{gathered}$$

Hence the series is convergent.