Question

Determine whether the following sequences diverge of converge?

Answer 1

1) {5, 0, 5, 0, 0, 5, 0, 0, 0} - diverges? i don't know how to explain though

Answer 2

2) sin(n) / (1+ sqrt{n}) This diverges as both the numerator and denominator approach infinity.

Answer 3

3) n^n / n!; diverges? I'm not so sure

Answer 4

I agree on your first answer
it looks like you have this pattern
5,0,5,0,0,5,0,0,0,5,0,0,0,0,5,0,0,0,0,0,5,...
basically it alternates between 5 and 0 forever and ever
so yeah totally agree with that one
for the second one sin(n) doesn't actually approach infinity
it stays between -1 and 1 forever and ever
\[\lim_{n \rightarrow \infty}\frac{\sin(n)}{1+\sqrt{n}} \text{ I think we can find this by squeeze theorem } \\ \frac{-1}{1+\sqrt{n} }\le \frac{\sin(n)}{1+\sqrt{n}} \le \frac{1}{1+\sqrt{n}}\]

Answer 5

OHHHH

Answer 6

find the limits for both -1/(1+sqrt(n)) and 1/(1+sqrt(n)) as n->infty

Answer 7

0

Answer 8

So that equals 0.

Answer 9

yep yep
so that means sin(n)/(1+sqrt(n))->0 as n gets bigger and bigger !

Answer 10

Amazing (:

Answer 11

What about the third part? :)

Answer 12

Which one do you think is bigger for large n
n^n or n!?

Answer 13

n^n hehehe

Answer 14

yah
so n^n/n! should diverge as n->infty

Answer 15

amazing :) Thank you

Answer 16

hmmm

Answer 17

hmmm?

Answer 18

nvm i misread it
thought someone said the first one converged

Answer 19

there is however a math way of saying it does not converge, it is basically negating the definition of convergence

Answer 20

how is sin{n} / 1 + sqrt{n} less than 1/ 1+ sqrt{n}?

Answer 21

would it not be bigger?

Answer 22

\[-1 \le \sin(x) \le 1 \]
first do you agree with this?

Answer 23

what is the biggest sine can be?

Answer 24

oh whoops

Answer 25

okay sounds great