Question

When do you use the root test vs ratio test to find the sum of a series?

Ex. Summation {n=1}^Infinity n!/(2^(n^2))

What's the best way to find the sum?

Thanks!

Answer 1

tests are tests .... they tend to be trial and error when trying to pick one

Answer 2

I've noticed most examples that require the ratio test have n! in them.. whereas most root tests have the denominator to a power of n or something of the like. I probably shouldn't pick up a pattern, but I'm trying to figure out the intuition behind these tests to make sure i use the best one most effectively.

Answer 3

Computations involving factorials are easier to do with the ratio test. Root test is primarily used if a given series contains something to the \(n\)th power. In my opinion, the ratio test is more often more helpful than the latter.

Answer 4

ah, i love your name

Answer 5

And for this particular series, I think a preliminary comparison, THEN the ratio/root test, would work best.

Thanks! :)

Answer 6

imnot sure if i recall a power equivalent of n! .... might be one out there .... gamma function comes to mind.

cant say i recall to much detail of a root test either ... to manyyears gone by i spose

ratio test:
\[\lim_{n\to ~inf}\frac{(n+1)!}{2^{(n+1)^2}}\frac{2^{n^2}}{n!}\]
\[\lim_{n\to ~inf}\frac{(n+1)\cancel{n!}}{2^{(n+1)^2}}\frac{2^{n^2}}{\cancel{n!}}\]
\[\lim_{n\to ~inf}\frac{(n+1)\cancel{n!}}{2^{n^2+2n+1}}\frac{2^{n^2}}{\cancel{n!}}\]
\[\lim_{n\to ~inf}\frac{(n+1)}{2^{2n+1}}\]

Answer 7

and yeah ok that makes sense.. i used the ratio test a problem that had a bunch of variables to the various powers/degrees of n and with that test i got a differnt answer from the root test.. is this something that should not have happened? are they supposed to eventually come to the same answer?

Answer 8

the summations should be the same yes, but working out a test for convergence may produce definitive results or not ....

Answer 9

when you factored out the n! in the second line.. where you wrote (n+1)(n!).. how did you know to do this? or better question.. can this be applied to any factorial.. like if i have (4n+9)! can i write that as (4n+9)(n!)?

Answer 10

i see, ok.. then i made a mistake somewhere with my ratio test.. :(

Answer 11

lets define k! as the product: k(k-1)(k-2)(k-3)...(3)(2)(1)

if we let k=n+1

(n+1)! = (n+1) (n) (n-1)(n-2)(n-3)...(3)(2)(1)
^^^^^^^^^^^^^^^^^^^^
this, by definition is n!

Answer 12

oki need to stare at this for a few minutes so itll sink in.. one min

Answer 13

if (4n+9)! lets see if we can get that to a similar structre

let k = 4n+9

(4n+9) (4n+8) (4n+7) ... (3)(2)(1)

i dont see that being a viable idea :/

Answer 14

ok i answered my own question while writing it out.. nvm i get it now haha

Answer 15

:) good job

Answer 16

patting myself on the back! wohoo.. idk why the k! threw me off.. but i just thought of it as a regular integer.. that you turn into a variable that equals n+1.. correct? and you find the value that way.. i get it but i can't explain it well.. in time

Answer 17

n^k = n(n-1)(n-2)....(n-k+1).. this is what wikipedia is telling me. now im failing to understand why the power is all of a sudden being throwin iat the end

Answer 18

we can consider it as how factorials are created:

1! = 1
2! = 2(1) = 2*1!
3! = 3(2)(1) = 3*2!
4! = 4(3)(2)(1) = 4*3!

Answer 19

n^k ? a number raised by an exponent, but the defintion you presented seems more like a permutation

Answer 20

there's a bar under the power k.. what does that mean?

Answer 21

so it's n^(k_) imagine the underscore is under the k

Answer 22

2^3 = 8

2(2-3+1) = 0 at best :)

Answer 23

this perhaps?
\[\large ^nP_k\]

Answer 24

@amistre64, op is referring to the second listed item: http://en.wikipedia.org/wiki/Factorial#Applications

It's just some weird notation the editors decided to use.