Question

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}}$

Short answer

Hence the given power series is convergent for all the values of$\mathit{x}$

Step-by-step solution

Given Information 

The power series is $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}}$

Definition of the interval of convergence.

The Interval of convergence is the interval in which the power series is convergent.

Find the interval.

The power series is$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}}$

Let ${\mathit{\rho }}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right|$.

Substitute the value of the power series in the formula above, the equation becomes as follows.

${\mathit{\rho }}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right|$

$$\begin{gathered} \mathbf{=}\left|\frac{\frac{x^{n+1}}{(2n+2)!}}{\frac{x^n}{\left(2n\right)!}}\right| \\ \mathbf{=}\left|\frac{x^{n+1}\left(2n\right)!}{(2n+2)!x^n}\right| \end{gathered}$$

Apply limits in the above equation.

$$\begin{gathered} \mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\left|\frac{x}{(2n+2)(2n+1)}\right| \\ \mathbf{=}\left|\frac{x}{\infty }\right| \\ \mathbf{=}\mathbf{0} \end{gathered}$$

The power series is convergent for $\mathit{\rho }\mathbf{<}\mathbf{1}$ .

Hence the given power series is convergent for all the values of $\mathit{x}$