Q9P - Mathematical Methods in Physical Sciences

Question

Test each of the following series for convergence.

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

Step-by-Step Solution

Answer

The series is convergent.

Step 1: Given Information

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

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

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.