Question

Can someone help me with this sequence problem?
http://dl.dropbox.com/u/6717478/Untitled.png

as far as I can tell it should converge at 0, but I don't know how to deal with the n! when evaluating the limit. Also, do you think that evaluating the limit is enough to answer the problem?

Answer 1

yeah it looks like it will converge because n! is greater than (-3)^n for large n
i am not sure how to differentiate n! either

Answer 2

I know that one need the ratio test a(n+1) / an

then the n! cancel and leave you with n+1

Answer 3

perhaps something about the Euler constant e

Answer 4

hmm I'm not sure about either of those methods

Answer 5

\[\sum_{n=0}^∞\frac{1}{n!}=e\]

Answer 6

so
\[\sum_{n=0}^∞ a_n=\sum_{n=0}^∞ \frac{(-3)^n}{n!}\]\[~~~~~~~~=e\sum_{n=0}^∞ (-3)^n\]

Answer 7

hmm i guess that does not suggest convergence though

Answer 8

n! always overpowers anything and everyting else for the most part

Answer 9

I get the answer that it diverges as the absolute value of a(n+1) / an >1

Answer 10

how about if I do something like this
3*( -1) <= 3* (-1)^n <= 3*( 1)
-------- -------- --------
n! n! n!

then the limit of the outer 2 go to 0 since the bottom goes to infinity. So the middle also goes to 0 by the Squeeze theorem.

Answer 11

\[\frac{3^n}{n!}=\frac{3}{1}\times \frac{3}{2}\times \frac{3}{3}\times ...\times \frac{3}{n}\]
\[<\frac{3}{1}\times \frac{3}{2}\times \frac{3}{3}\times \frac{3}{4}\times \frac{3}{4}\times...\times\frac{3}{4}\]

Answer 12

\[=\frac{3}{1}\times \frac{3}{2}\times \frac{3}{3}\times (\frac{3}{4})^{n-3}\]

Answer 13

take the limit as n goes to infinty and you get zero

Answer 14

3^(n+1) n! 3^n * 3 1
------- * ----- = ------ * -----
(n+1)! 3^n n+1 3^n

Answer 15

gets me to 3/n+1 ahhh and as n --> infinity and beyond = 0

so when I dont forget to carry a term 0 < 1 and it converges by ratio test

Answer 16

Your method is still a little confusing to me satellite, as far as I can tell my squeeze theorem way gives me the answer I'm looking for though

I don't think I've learned the ratio test yet kantalope

Answer 17

@kantalope you are applying the ratio test, for summing a series, not for computing a limit of the sequence. then again if series converges, certainly the limit of the sequence is 0

Answer 18

ratio test is not for limit of a sequence

Answer 19

problem with your squeeze theorem is that you are ignoring the fact that 3 is being raised to the power of n

Answer 20

figuring out which test is harder than the tests themselves

Answer 21

it is not
\[3(-1)^n\] is is
\[(-3)^n\] which is entirely different

Answer 22

hmm you're right

Answer 23

all i did was replace every number in the denominator larger than 4 by 4 and so i get the inequality
\[\frac{3^n}{n!}<\frac{9}{2}\times (\frac{3}{4})^{n-3}\] and since
\[\frac{3}{4}<1\] this goes to zero as n goes to infinity

Answer 24

but how does that show that original equation goes to 0 as n goes to infinity?

Answer 25

because
q

Answer 26

because
a) each term is positive and
b) it is less that something that goes to zero

Answer 27

so since it is an increasing sequence and something that is greater than it goes to 0 then it also goes to 0?

Answer 28

how come your solution doesn't take (-3)^n but only 3^n?