Question

Determine the convergence or divergence of the series:

Answer 1

\[1 + \frac{ 1*3 }{ 1*2*3 }+ \frac{1*3*5}{1*2*3*4*5} + \frac{1*3*5*7}{1*2*3*4*5*6*7} ...\]

Answer 2

Basically Im just not sure how to put it into series form. I figured something like (2n-1)/(2n-1)(2n), since it seems like the denominator needs two such terms. And yes, I know what I typed cancels. But I cannot find a way to have the denominator maintain an odd term and an even term and the numerator have an odd term without some sort of mishap. The main mishap is that I do not know how to get the first term to be a 1 and still maintain the pattern of the rest of the series. If the first 1 wasn't there it'd be no issue -_-

Answer 3

if you simplify it you just get\[1+\frac12+\frac1{2\cdot4}+\frac1{2\cdot4\cdot6}+\cdots\]so you just need to find a way to represent the denominators as factorials
(hint: think about using an exponential factor outside the factorial term to get the pattern)

Answer 4

another hint: imagine taking a factor of two out of the denominator

Answer 5

Im looking at (2n)! on bottom but with something on top cancelling out the odd numbers. But.....

Answer 6

consider the sequence:\[1=2^0=1!\\2=2^1=2!\\2*4=2^2(1*2)=2^2(2!)\\2*4*6=2^3(1*2*3)=2^3(3!)\]seeing a pattern yet? :)

Answer 7

Yeah, I see it. Sorry, was busy for a sec, not to leave ya sitting there.

Answer 8

better illustrated like this:\[\large1=2^0=2^0(0!)\\\large2=2^1=2^1(1!)\\\large2*4=2^2(1*2)=2^2(2!)\\\large2*4*6=2^3(1*2*3)=2^3(3!)\]so in terms of n...?

Answer 9

Right, 2^n*(n!). I guess Im just looking at it still, something is bugging me even though I see it, lol.

Answer 10

well keep staring, that's a good way to learn :)
you are right though, we can write this as\[\sum_{n=0}^\infty\frac1{2^n(n!)}\]

Answer 11

Alrighty, nvm, I got it. So what are you looking for when you first see that? Like, I can think 2n for the bottom and then I just wonder how to avoid 0. I guess exponentials and factorials avoid 0 in the bottom when n = 0. If you even can explain it, doyou have a thought process with this or is it just experience?

Answer 12

Actually I have to admit having had this same issue once before, and I just happened to remember that factoring out the 2's made itunderstandable in terms of n, so it's a matter of experience I guess :p

Answer 13

I think a lot of math is like that; you don't memorize the exact rule, but you go "oh, I think that factor of 2 was they key to fixing it...|

Answer 14

my first thought was to start cancelling though :)

Answer 15

Right. Fair enough, lol. I was cancelling at first, too, but I was running into similarproblems. And then I was worried that too much simplifying might cause me to not see an answer as opposedto making it more clear. I wasn't willing to giveup (2n-1) on top just yet I guess.

Answer 16

yeah, you do need to do that cancelling trick for the odd numbers actually (you cancel with 2n+1 actually), but for the evens you just factor out 2's all the way down

Answer 17

Yeah. I think there are going to end up being some recognition things as well as little rules to pick up on, though. Like treating cos(n) as \[\sum_{n=1}^{\infty}(-1)^{n}\]Oh well, as I see more of these they don't seem to be that bad. Thanks :)

Answer 18

very welcome :)