Question

checking for diverging and converging:
sum: n=1 -> infinity

sin 4n / 4^2 I'm gonna try ratio test a(n+1) / a(n)

I think I'm on the right path and get to
(sin 4(n+1) / 4) * (1 / sin 4n)

I'm not sure how to cancel the sin parts??

Answer 1

\[\sum_{1}^{\infty} (sin 4n) /(4^{n})\]

\[((\sin 4(n+1))/4) * (1/(\sin 4n)\]

Answer 2

might be better to try an integral test ?

Answer 3

just a thought, prolly wrong tho

Answer 4

aint graphed it yet but im thinking it goes like this

Answer 5

http://www.wolframalpha.com/input/?i=sin%284n%29%2F4%5En

yep

Answer 6

sin(4(n+1)) and sin(4n) for large values of n are for all practical purposes the same

Answer 7

and im reading it wrong arent i

thats 4^2 underneath or 4^n?

Answer 8

oh yeah I goofed up the first part its 4^n in the denominator ---

is there any manipulation I can do with the sin 4(n+1) / sin 4n or can i jump straight to that is basically a limit of 1 at infinity?

Answer 9

\[lim\ \frac{sin(4n)}{4^n}*\frac{4^{n-1}}{sin(4n-4) }\]
\[lim\ \frac{sin(4n)}{4}*\frac{1}{sin(4n-4)}\]
\[\frac 14\ lim\ \frac{sin(4n)}{sin(4n-4)}\]
might wanna see if you can split this up with the sin(a+b) identities

Answer 10

http://www.wolframalpha.com/input/?i=lim+sin%284n%29%2F4%5En+to+inf

or do Lhopital Rule

Answer 11

sin(4n) doesnt for to inf or 0 tho so Lhop might be out as far as proofing

Answer 12

http://www.wolframalpha.com/input/?i=lim+sin%284n%29%2Fsin%284n%2B4%29+to+inf

i loathe limits lol

Answer 13

ratio aint gonna work for you on this

Answer 14

the limit for sin at infinity would be one in any case, ya?

which makes sense with the wolfram answer 1/really big number = 0

Answer 15

i dont know what the wolfs basing it on; but i know that the sin will occilate back and forth inbetween 1/4^n which goes to zero

Answer 16

4n^(-1) is the amplitude of sin; and as n goes larg; 4^n goes small and approaches zero

Answer 17

the squeeze thrm would say that sin gets squaeezed to zero along with it