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I need help with this practice problem It asks to solve (a) and (b) Please put these separately so I know which is which ^

I Need Help With This Practice Problem It Asks To Solve A And B Please Put These Separately So I Know Which Is Which class=

Sagot :

Given the series

[tex]\sum (\frac{2n!}{2^{2n}})[/tex]

Let

[tex]a_n=\frac{2n!}{2^{2n}}[/tex]

(a) Find r using the using the formula

[tex]r=\lim _{n\to\infty}\frac{a_{n+1}}{a_n}[/tex]

Substitute the given values.

[tex]\begin{gathered} r=\lim _{n\to\infty}\frac{2(n+1)!}{2^{2(n+1)}}\cdot\frac{2^{2n}}{2n!} \\ =\lim _{n\to\infty}\frac{(n+1)}{2^2}\frac{^{}}{} \\ =\infty \end{gathered}[/tex]

(b) By Ratio test, if the value of r is greater than 1, then the series is divergent. Here r > 1. So, the series is divergent.