Question

need help with this series problem

Answer 1

\[\sum_{n=1}^{\infty}\frac{ 1*3*5.....*(2n-1) }{ 2*5*8....*(3n-1) }\]

Answer 2

Hey I was just looking at your past questions and noticed you left a reply I didn't reply to.

Answer 3

Are you still needing help on that one too?

Answer 4

yeah

Answer 5

ok first thing is this I guess
what is the question here

Answer 6

but the thing i dont get on this is why it has (2n+1)/(3n+2)

Answer 7

why all of that simplified to that right?

Answer 8

yeah

Answer 9

\[\frac{(2n+1)! (3n-1)!}{(3n+2)!(2n-1)!}\]
you started with this right?

Answer 10

no how you get that

Answer 11

\[x!=x(x-1)(x-2)(x-3)(x-4) \cdots (3)(2)(1)\]

Answer 12

other example
\[(2x-1)!=(2x-1)(2x-2)(2x-3)(2x-4) \cdots (2x-x)(2x-[x-1]) ! \\ (2x-[x-1])!=(2x-x+1)!=(x+1)!=(x+1)(x)(x-1) \cdots (3)(2)(1)\]

Answer 13

anyways
\[(2n+1)!=(2n+1)(2n)(2n-1)!\]
\[(3n+2)!=(3n+2)(3n+1)(3n)(3n-1)!\]

Answer 14

you will get to cancel some factorials factors

Answer 15

\[\frac{(2n+1)! (3n-1)!}{(3n+2)!(2n-1)!} \\ \frac{(2n+1)(2n)(2n-1)! \cdot (3n-1)!}{(3n+2)(3n+1)(3n)(3n-1)! (2n-1)!}\]

Answer 16

but the problem doesnt have factorials

Answer 17

\[\frac{(2n+1)! (3n-1)!}{(3n+2)!(2n-1)!} \\ \frac{(2n+1)(2n)\cancel{(2n-1)!} \cdot \cancel{(3n-1)!}}{(3n+2)(3n+1)(3n)\cancel{(3n-1)! } \cancel{(2n-1)!} } \\ \frac{2}{3} \frac{2n+1}{(3n+2)(3n+1)}\]

Answer 18

which is a different looking then what they got

Answer 19

writing it as factorials helps me simplify it

Answer 20

i just noticed that he used ratio test and plugged in n+1

Answer 21

is the 2n part of the factorial as well?

Answer 22

(2n+1)!=(2n+1)(2n+0)(2n-1)!
well if you are asking if is a factor of (2n+1)! yes

and

(3n+2)!=(3n+2)(3n+1)(3n+0)(3n-1)!
so 3n is a factor of (3n+2)!

Answer 23

i think i got it now thanks for the help! :)

Answer 24

http://tutorial.math.lamar.edu/problems/calcii/alternatingseries.aspx
I don't see number 4 here?

Answer 25

if you could answer the previous question please

Answer 26

you just gotta scroll down

Answer 27

oh sorry
i'm an idiot

Answer 28

its okay

Answer 29

what about them finding f'
do you understand that part?

Answer 30

do you know why they are finding f'?

Answer 31

yeah i get how to do that

Answer 32

and you know the why part too

Answer 33

f' tells us if the function is decreasing or increasing

Answer 34

f'=0 tells us where it does neither
f' dne cancel also tell us that too depending on if the function is continuous there

Answer 35

or i guess say it does neither there :p

Answer 36

oh no i dont understand that part very much

Answer 37

f'=0
when the top=0

Answer 38

9-x^2=0
when x=?

Answer 39

x = 3

Answer 40

and ? or I should rather say or ?

Answer 41

it increases cause its positive?

Answer 42

no I was asking for another solution
9-x^2=0 when x=3 or x=-3
f' exist everywhere here since the denominator is never 0

but anyways f'=0 tells us the function is resting
it is resting at x=3 or x=-3

Answer 43

we are rough picture for fun