Q6P - Mathematical Methods in Physical Sciences

Question

Test each of the following series for convergence.

$\underset{\mathbf{ }}{\mathbf{\sum }{\displaystyle \frac{\mathbf{1}\mathbf{+}\mathbf{i}}{{\mathbf{n}}^{\mathbf{2}}}}}$ 

Step-by-Step Solution

Answer

 The series is convergent.

Step 1: Given Information

The series is  $\underset{ }{\sum {\displaystyle \frac{1-i}{n^2}}}$.

Step 2: Definition of the Convergent series.

A series can be categorized to be convergent if the terms of a series propagate to zero when the number of terms moves towards infinity.

Step 3: Test the convergence.

The series is $\sum _{ }\frac{1+\mathrm{i}}{{\mathrm{n}}^2}$.

 $$\begin{gathered} {\mathrm{A}}_{\mathrm{n}}=\frac{1+\mathrm{i}}{{\mathrm{n}}^2} \\ {\mathrm{A}}_{\mathrm{n}+1}=\frac{1+\mathrm{i}}{{\left(\mathrm{n}+1\right)}^2} \end{gathered}$$

 

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

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

Test fails.

 

Find the sum of the series.

$$\begin{gathered} \mathrm{S}={\int }_0^{\infty }\left(\frac{1+\mathrm{i}}{{\mathrm{n}}^2}\right)\mathrm{dn} \\ ={\left[\frac{-\left(1+\mathrm{i}\right)}{\mathrm{n}}\right]}_1^{\infty } \\ =1+\mathrm{i} \\ \mathrm{S}=\sqrt{2} \end{gathered}$$ 

 

The sum is finite.

 

Hence the series is convergent.