Respuesta :

Given series,

[tex]\sum ^{\infty}_{n\mathop=1}(\frac{2n!}{2^{2n}})[/tex]

a.The value of r is

[tex]\begin{gathered} r=\lim _{n\to\infty}|\frac{a_{n+1}}{a_n}| \\ \Rightarrow r=\lim _{n\to\infty}\frac{(2n+1)!}{2^{2(n+1)}}\times\frac{2^{2n}}{(2n)!} \\ \Rightarrow r=\lim _{n\to\infty}\frac{2(n+1)}{4}=\infty \end{gathered}[/tex]

b. Since r is infinity so the series is divergent.

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