Question

For any of you calc II fans: Im stuck on this!

sum_{n=3}^{infty} (n^3)/(2n+1)!
I believe this converges, but using a ratio test, it appears to be infinity over infinity, which is inconclusive. Im doing something wrong. Please help me out someone!

Answer 1

what do u get after simplifying the ratio ?

Answer 2

I get 2n+1/2n+2, which applying a lim as x goes to infinity would be inconclusive, infinity over infinity.

Answer 3

\[\rm \begin{align} \lim\limits_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3/(2(n+1)+1)!}{n^3/(2n+1)!}\right| \\~\\
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3/(2n+3)!}{n^3/(2n+1)!}\right| \\~\\


\end{align}\]

Answer 4

Can I pull out n! from (2n+1)! or would it be 2n! ? I am not very familiar with the rules of factorials.

Answer 5

\[n! = n\cdot (n-1)!\]

Answer 6

\[n! = n\cdot (n-1)! = n\cdot(n-1)\cdot (n-2)!\]

Answer 7

you can keep going like that till you reach 1

Answer 8

lets rearrange the fractions first

\[\rm \begin{align} \lim\limits_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3/(2(n+1)+1)!}{n^3/(2n+1)!}\right| \\~\\
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3/(2n+3)!}{n^3/(2n+1)!}\right| \\~\\
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3(2n+1)!}{n^3(2n+3)!}\right| \\~\\


\end{align}\]

Answer 9

I just tried this on paper, and I end up with \[(2n+1)!/(2n+3)!\]

Answer 10

can i pull out a factorial n! from these?

Answer 11

you cannot cancel (n+1)^3 and n^3

Answer 12

I still do not understand how to get anywhere with this. I think you have the correct fraction, I can arrive at that as well, however I don't see how to obtain the ratio from it.

Answer 13

lets split (2n+3)! as (2n+3)(2n+2)(2n+1)!
\[\rm \begin{align} \lim\limits_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3/(2(n+1)+1)!}{n^3/(2n+1)!}\right| \\~\\
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3/(2n+3)!}{n^3/(2n+1)!}\right| \\~\\
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3(2n+1)!}{n^3\color{red}{(2n+3)!}}\right| \\~\\
& = \lim\limits_{n\to \infty}\left|\dfrac{(n+1)^3(2n+1)!}{n^3\color{Red}{(2n+3)(2n+2)(2n+1)!}}\right| \\~\\


\end{align}\]

Answer 14

cancel the (2n+1)! top and bottom

Answer 15

ok, now I am at \[(n+1)^3/(n^3*(2n+3)(2n+2))\]

Answer 16

yes take the limit

Answer 17

this is still infinity over infinity

Answer 18

notice that degree of numerator = 3
degree of denominator = 4

Answer 19

the higher degree denominator clearly pulls the value of fraction to 0 as n -> infinity

Answer 20

I get 5 as the degree in the denominator. I foiled the (2n+3)(2n+2) which is 4n^2+10n+6 and distributed n^3 to have n^5 as the highest degree. Am I doing this wrong as well?

Answer 21

Oh right, degree of denominator is 5

Answer 22

ok, I know this doesn't change it. So, the faster approach of the denominator makes this become zero.

Answer 23

Thank you for your help ganeshie8