91Ó°ÊÓ

The given series is, $\underset{n-0}{\overset{∞}{∑}}{\left(-1\right)}^n{z}^{2n}\underset{k-0}{\overset{n}{∑}}\left(\frac{1}{\left(2k+i\right)!}\right)$.

The ratio test is given as:

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

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

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

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

The ratio test is given as follows:

$$\begin{gathered} {\mathrm{ÒÏ}}_n=\left|\frac{a_{n+1}}{a_n}\right| \\ \mathrm{ÒÏ}=\underset{n→∞}{lim}{\mathrm{ÒÏ}}_n \\ \mathrm{ÒÏ}=\underset{n→∞}{lim}\left|\frac{a_{n+1}}{a_n}\right| \end{gathered}$$

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

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

From equation (1) as follows:

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

Since, we know that:

$$\begin{gathered} \mathrm{ÒÏ}=\underset{\mathrm{n}→∞}{\lim }{\mathrm{ÒÏ}}_n \\ \mathrm{ÒÏ}=\underset{\mathrm{n}→∞}{\lim }\left|\frac{a_{n+1}}{a_n}\right| \end{gathered}$$

Therefore, we can write as follows:

$$\begin{gathered} \mathrm{ÒÏ}=\underset{\mathrm{n}→∞}{\lim }{\mathrm{ÒÏ}}_n \\ \mathrm{ÒÏ}=\underset{\mathrm{n}→∞}{\lim }\frac{{\left(-1\right)}^{n+1}{\displaystyle \frac{1}{\left(2n+2\right)!}}z\left(2n+2\right)}{{\left(-1\right)}^{n+1}\frac{1}{\left(2n+1\right)!}z\left(2n+2\right)} \\ \mathrm{ÒÏ}=\underset{n→∞}{lim}\frac{z^{\left(2n\right)}{\left(-1\right)}^n\left(-1\right)z^2}{\left(2n+2\right)!}×\frac{\left(2n+1\right)}{z^{2n}{\left(-1\right)}^n} \\ \mathrm{ÒÏ}=\underset{n→∞}{lim}\frac{\left(-1\right)z^2}{\left(2n+2\right)\left(2n+1\right)!}×\left(2n+1\right)! \end{gathered}$$

Solve the factorial in the above equation as follows:

$$\begin{gathered} \mathrm{ÒÏ}=\underset{n→∞}{lim}\frac{\left(-1\right)z^{\left(2\right)}}{\left(2n+2\right)} \\ \mathrm{ÒÏ}=\underset{n→∞}{lim}\left(\frac{1}{{\displaystyle \frac{1}{n}}+i}\right) \end{gathered}$$

By apply limit $n→∞$ in the above equation, obtain:

$$\begin{gathered} \mathrm{ÒÏ}=\underset{n→∞}{lim}{\mathrm{ÒÏ}}_n \\ \mathrm{ÒÏ}=0 \end{gathered}$$