Q11P - Mathematical Methods in Physical Sciences

Question

Test each of the following series for convergence.

 $\mathbf{\sum }_{\mathbf{ }}{\left(\frac{2+i}{3-4i}\right)}^{\mathbf{2}\mathbf{n}}$

Step-by-Step Solution

Answer

The series converges, i.e., $\rho <1$ .

Step 1: Given Information.

The given series, i.e. $\sum _{ }{\left(\frac{2+\mathrm{i}}{3-4\mathrm{i}}\right)}^{2\mathrm{n}}$, 

Step 2: Definition of Convergent and Divergent series.

A convergent series is one in which the partial sums all gravitate to the same finite number, also known as a limit. Divergent refers to any series that is not converging.

Step 3: Calculate the value of &#961; n .

Find the value of  ${\mathrm{a}}_{\mathrm{n}}$ and ${\mathrm{a}}_{\mathrm{n}+1}$

$$\begin{gathered} {\mathrm{a}}_{\mathrm{n}}={\left(\frac{2+\mathrm{i}}{3-4\mathrm{i}}\right)}^{2\mathrm{n}} ...\left(1\right) \\ {\mathrm{a}}_{\mathrm{n}+1}={\left(\frac{2+\mathrm{i}}{3-4\mathrm{i}}\right)}^{2(\mathrm{n}+1)} \\ {\mathrm{a}}_{\mathrm{n}+1}={\left(\frac{2+\mathrm{i}}{3-4\mathrm{i}}\right)}^{2\mathrm{n}+2} ...\left(2\right) \end{gathered}$$ 

 

 

If $\mathrm{\rho }<1$ series converges, if $\mathrm{\rho }>1$ then diverges.

$\mathrm{\rho }=\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{\rho }}_{\mathrm{n}}$ 

Step 4: Test the series for convergence.

Use the ratio test.

$$\begin{gathered} {\mathrm{\rho }}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}} \\ ={\left(\frac{2+\mathrm{i}}{3-4\mathrm{i}}\right)}^{2\mathrm{n}+2-2\mathrm{n}} \\ ={\left(\frac{2+\mathrm{i}}{3-4\mathrm{i}}\right)}^2 \\ =-0.187+0.07\mathrm{i} \end{gathered}$$

Calculate the value of $\mathrm{\rho }$ , i.e.,

$$\begin{gathered} \mathrm{\rho }=\underset{\mathrm{n}\to \infty }{\lim }\left|{\mathrm{\rho }}_{\mathrm{n}}\right| \\ =\underset{\mathrm{n}\to \infty }{\lim }\left|-0.187+0.07\mathrm{i}\right| \\ \mathrm{\rho }=0.2 \end{gathered}$$

Hence, the series converges, i.e., $\rho <1$ .