Question

another series question -- my answer disagrees with the ans in my solutions manual, and I agree with their solution but can't find the fault in mine so I'm here to look for someone to pick at my mistakes... =) will take time to throw down my sol'n

Answer 1

\[\sum_{n=1}^{\infty}\frac{ 2\times4\times6...\times(2n) }{ n! }\]

Answer 2

and I'm asked if it converges absolutely, conditionally or diverges, so I thought I'd just use the ratio test

Answer 3

but after working through it all I end up with 1/n

Answer 4

are you really sure the denominator is n! and not n!^2 ?

Answer 5

\[|\frac{ 2\times4\times6...\times(2(n+1)) }{ (n+1)! } \times \frac{ n! }{ 2\times4\times6...\times(2n) }|\]
and then I simplified...

yeah, it's definitely just n!

Answer 6

so when simplifying, is it cool to just cancel all of the 2 times 4 times 6 and keep just the 2(n+1) and 2(n)? That may be where I'm slipping up

Answer 7

that simplifies to 2(n+1)/(n+1) which is same as 2 right ?

Answer 8

I simplified to 1/n =o I'll try a third time, then, it's probably just me being a dunce

Answer 9

first simplify the factorial terms maybe :


n!/(n+1)! = 1/(n+1)

Answer 10

\[|\frac{ (2(n+1)) }{ (n+1)! } \times \frac{ n! }{ (2n) }|\]\[|\frac{ (n+1) }{ (n+1)! } \times \frac{ n! }{ (n) }|\]\[\frac{ (n+1) }{ (n+1)(n!) } \times \frac{ n! }{ (n) }\]

Answer 11

no wait, there is a mistake in first line

Answer 12

am I messing with the factorials wrong? Totally new to them, also a little slow, so it's entirely possible lol

Answer 13

that (2n) should not be there in the denominator

Answer 14

But when you cancel all the product of even integers, doesn't the 2n remain? D=

Answer 15

let me show u step by step. start with \[\left|\frac{ 2\times4\times6...\times(2(n+1)) }{ (n+1)! } \times \frac{ n! }{ 2\times4\times6...\times(2n) }\right| \]

Answer 16

same as

\[\left|\frac{ 2\times4\times6...\times(2n)\times (2(n+1)) }{ (n+1)! } \times \frac{ n! }{ 2\times4\times6...\times(2n) }\right| \]

yes ?

Answer 17

ahh, okay so then there's also a 2n if we're going up until 2(n+1)

Answer 18

thats it!

Answer 19

Thank you!!!!!!!!!! You've helped me a bunch tonight I hope I can repay the favour sometime

Answer 20

lol that reminds me of godfather movie somehow