Answer to Question #161109 in Calculus for vadris

Question

Consider the infinite series defined by $∞∑n=1 ((2n!)/(2^2n))$.

(a) What is the value of r from the ratio test?

(b) What does this r value tell you about the series?

Answer 1

Here we have the series $S_n$ as,

$S_n= \displaystyle\sum_{n=1}^\infin\frac{(2n)!}{2^nn}$

(a)

Now, let $a_n=$ $\frac{(2n)!}{2^nn}$

So, $a_{n+1}=\frac{(2n+2)!}{2^{n+1}(n+1)}$

So, $r=\frac{a_{n+1}}{a_n}$

$r=\frac{a_{n+1}}{a_n}\\~\\ \Rightarrow r=\frac{(2n+2)!\cdot n\cdot 2^n}{2^{n+1}\cdot(n+1)\cdot(2n)!}\\~\\ \Rightarrow r=n(n+1)$

Therefore, $r=n(n+1)$

(b)

Now, we will evaluate $\lim\limits_{n\to\infin}r$

So,

$\lim\limits_{n\to\infin}r\\ =\lim\limits_{n\to\infin}n(n+1)=\infin$

So, as $r$ tends to $\infin$ ,we can conclude that the series $S_n$ diverges.