Answer to Question #161109 in Calculus for vadris

Question #161109

Consider the infinite series defined by n=1((2n!)/(22n))∞∑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?


1
Expert's answer
2021-02-23T06:26:49-0500

Here we have the series SnS_n as,

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


(a)


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


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


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



r=an+1an r=(2n+2)!n2n2n+1(n+1)(2n)! r=n(n+1)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)r=n(n+1)


(b)


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


So,


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

So, as rr tends to \infin ,we can conclude that the series SnS_n diverges.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment