Question

Calculus. USING the ratio test to determine convergence or divergence

Answer 1

\[\sum_{n=1}^{\infty}(-1)^n \frac{ 3^n(n!) }{ 7*12*17*...*(5n+2) }\]

Answer 2

was hoping someone could walk me through this. I am really struggling with making the ratio test out of this

Answer 3

does \[a_n=\frac{ (-1)^n 3^n(n!) }{ (5n+2) }\]

Answer 4

\[a_n=\frac{(-1)^n3^n(n!)}{(5n+2)!}\]

Answer 5

oh.... ok... so then...

Answer 6

find a_(n+1)

Answer 7

that should be a pretty fast task
you replace old n with (n+1)
have you done that?

Answer 8

sorry, my signal is crap, im trying to put it in the equation thing

Answer 9

\[a_n+1= \frac{ (-1)^{n+1}3^{n+1}(n+1)! }{ (5n+3) }\]

Answer 10

\[\lim_{n \rightarrow \infty}| \frac{a_{n+1}}{a_n} | \\ =\lim_{n \rightarrow \infty}| \frac{(-1)^{n+1}3^{n+1}(n+1)!}{(5n+7)!} \cdot \frac{(5n+2)!}{(-1)^n3^n(n)!}|\]

Answer 11

if you have
f(n)=(5n+2)!
then
f(n+1)=(5[n+1]+2)!=(5n+5+2)!=(5n+7)!

Answer 12

ok, thanks, that was one thing i knew i was really doing wrong.... let me see what i have here

Answer 13

so, the ones and the 3's cancel, right?

Answer 14

so i would look at it in separate parts for now and then put it back together
like I would "simplify" the following:
\[\frac{(-1)^{n+1}}{(-1)^{n}}=? \\ \frac{3^{n+1}}{3^n}=? \\ \frac{(n+1)!}{n!} =? \\ \frac{(5n+2)!}{(5n+7)!}=?\]

Answer 15

because n+1-n = 1

Answer 16

yes so for that one you will have 3^1 or just 3

Answer 17

(n+1)! = (n+1) (n)! , right?

Answer 18

yep
\[\frac{(-1)^{n+1}}{(-1)^{n}}=? \\ \frac{3^{n+1}}{3^n}=3^{n+1-n}=3^1=3 \\ \frac{(n+1)!}{n!} =\frac{(n+1) \cdot n!}{n!}=? \\ \frac{(5n+2)!}{(5n+7)!}=? \]

Answer 19

first, why doesnt the 3's cancel since there is a 3 in numer and denom?

Answer 20

and im really not sure how to do the last one...honestly

Answer 21

Are you referring to the our application of exponents? \[\frac{3 \cdot 3 \cdot 3}{3 \cdot 3}=?\]
do you know what these equals ?

Answer 22

ooohhh ok

Answer 23

that one would equal 3
like because not all the 3's cancel

Answer 24

i was looking at that wrong.... so, how do i do the last part? I'm really not sure

Answer 25

\[(5n+7)!=(5n+7) \cdot (5n+6) \cdot (5n+5) \cdot (5n+4) \cdot (5n+3) \cdot (5n+2)!\]

Answer 26

just like before when you did (n+1)!=(n+1) times (n+1-1)!

Answer 27

for example 5!=(5)(4)(3)(2)(1) or 5(5-1)(4-1)(3-1)(2-1)
you are taking one away from the previous one

Answer 28

so, (5n+2)! cancels and we are left with the other ones in the denom?

Answer 29

yes
which would mean that our deg on bottom will be much bigger than the deg on top
since we have
\[\lim_{n \rightarrow \infty}|\frac{(-1)(3)(n+1)}{(5n+7)(5n+6)(5n+5)(5n+4)(5n+3)}| \]

Answer 30

and so therefore you know the limit is...

Answer 31

is it \[\frac{ \infty }{ \infty }\]?

Answer 32

I was looking for L=0

Answer 33

ok, because the denominator is larger than the numer?

Answer 34

if you have a polynomial over a polynomial while n goes to infinity and the deg on bottom is larger than the degree on top then the limit is 0

Answer 35

wow... i thank you sooooo much! you have no idea how much this is gonna help me

Answer 36

if you have a polynomial over a polynomial while n goes to infinity and the deg on top is larger than the degree on bottom then the limit is one of the infinitys

Answer 37

i need to see thngs worked out fully like this, or i just dont see it

Answer 38

Anyways since we have L=0 which menas L is less than one that means the series is ....

Answer 39

\[L >1=\] divergent?

Answer 40

except 0 is not greater than one

Answer 41

ok so \[L <1\] is convergent

Answer 42

yes L is less than 1 so the series is convergent

Answer 43

thank you soo much @myininaya

Answer 44

np