Q12P - Mathematical Methods in Physical Sciences

Question

Test each of the following series for convergence.

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

Step-by-Step Solution

Answer

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

Step 1: Given Information.

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

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  ${\mathrm{a}}_{\mathrm{n}+1}$

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

 

If $\mathrm{\rho }<1$ series converges, if $\rho >1$ 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 and put  $\left(\mathrm{n}+1\right)!=\left(\mathrm{n}+1\right)\left(\mathrm{n}!\right)$

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

Calculate the value of $\rho$ , i.e.,

$$\begin{gathered} \mathrm{\rho }=\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{3+2\mathrm{i}}{\mathrm{n}+1}\right| \\ =\left|\frac{3+2\mathrm{i}}{\infty }\right| \\ =0 \end{gathered}$$

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