Q17P - Mathematical Methods in Physical Sciences

Question

Verify the series in (7.3) by computer. Also show that it can be written in the form $\mathbf{\sum }_{\mathbf{n}\mathbf{-}\mathbf{0}}^{\mathbf{\infty }}{\left(-1\right)}^{\mathbf{n}}{\mathit{z}}^{\mathbf{2}\mathbf{n}}\mathbf{\sum }_{\mathbf{k}\mathbf{-}\mathbf{0}}^{\mathbf{n}}\frac{\mathbf{1}}{\left(2k+1\right)\mathbf{!}}$. Use this form to show by ratio test that the series converges in the disk $\left|z\right|\mathbf{<}\mathbf{1}$.

Step-by-Step Solution

Answer

The series for convergence is $\rho =\underset{n\to \infty }{lim}{\rho }_n=0$.

Step 1: Given data

The given series is, $\sum _{n-0}^{\infty }{\left(-1\right)}^n{z}^{2n}\sum _{k-0}^n\left(\frac{1}{\left(2k+i\right)!}\right)$ .

Step 2: Concept of Ratio test

The ratio test is given as:

 $$\begin{gathered} {\mathit{\rho }}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right| \\ \mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{\rho }}_{\mathbf{n}} \\ \mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\left|\frac{a_{n+1}}{a_n}\right| \end{gathered}$$

A geometric series converges if $\left|r\right|\mathbf{<}\mathbf{1}$.

A geometric series diverges if $\left|r\right|\mathbf{>}\mathbf{1}$.

Step 3: Solve to find the ration test of the given series

Let the series be, $\sum _{ }\left(\frac{1}{n^2}+\frac{i}{n}\right)$ .                                                               …… (1)

The ratio test is given as follows:

 $$\begin{gathered} {\rho }_n=\left|\frac{a_{n+1}}{a_n}\right| \\ \rho =\underset{n\to \infty }{lim}{\rho }_n \\ \rho =\underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right| \end{gathered}$$

A geometric series converges if $\left|r\right|<1$.

A geometric series diverges if $\left|r\right|>1$.

From equation (1) as follows:

 $$\begin{gathered} a_n=\sum _{n-0}^{\infty }{\left(-1\right)}^n{z}^{2n}\sum _{k-0}^n\left(\frac{1}{\left(2k+1\right)!}\right) \\ {a}_n=\sum _{n-0}^{\infty }{\left(-1\right)}^n\frac{1}{\left(2n+1\right)!}{z}^{2n} \\ {a}_{n+1}=\sum _{n-0}^{\infty }{\left(-1\right)}^{n+1}\frac{1}{\left(2n+1\right)+1)!}{z}^{2(n+1)} \\ {a}_{n+1}=\sum _{n-0}^{\infty }{\left(-1\right)}^{n+1}\frac{1}{\left(2n+1\right)!}{z}^{\left(2n+2\right)} \end{gathered}$$

Step 4: Calculation for the series of convergence

Since, we know that:

 $$\begin{gathered} \rho =\underset{\mathrm{n}\to \infty }{\lim }{\rho }_n \\ \rho =\underset{\mathrm{n}\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right| \end{gathered}$$

Therefore, we can write as follows:

 $$\begin{gathered} \rho =\underset{\mathrm{n}\to \infty }{\lim }{\rho }_n \\ \rho =\underset{\mathrm{n}\to \infty }{\lim }\frac{{\left(-1\right)}^{n+1}{\displaystyle \frac{1}{\left(2n+2\right)!}}z\left(2n+2\right)}{{\left(-1\right)}^{n+1}\frac{1}{\left(2n+1\right)!}z\left(2n+2\right)} \\ \rho =\underset{n\to \infty }{lim}\frac{z^{\left(2n\right)}{\left(-1\right)}^n\left(-1\right)z^2}{\left(2n+2\right)!}\times \frac{\left(2n+1\right)}{z^{2n}{\left(-1\right)}^n} \\ \rho =\underset{n\to \infty }{lim}\frac{\left(-1\right)z^2}{\left(2n+2\right)\left(2n+1\right)!}\times \left(2n+1\right)! \end{gathered}$$

Solve the factorial in the above equation as follows:

 $$\begin{gathered} \rho =\underset{n\to \infty }{lim}\frac{\left(-1\right)z^{\left(2\right)}}{\left(2n+2\right)} \\ \rho =\underset{n\to \infty }{lim}\left(\frac{1}{{\displaystyle \frac{1}{n}}+i}\right) \end{gathered}$$

By apply limit $n\to \infty$  in the above equation, obtain:

 $$\begin{gathered} \rho =\underset{n\to \infty }{lim}{\rho }_n \\ \rho =0 \end{gathered}$$