Question

Determine whether the sequence converges or diverges. If it converges, find the limit.

A sub n = sin2n/1+sqrt n

Answer 1

is that a sin(2n), or a sin^2 (n)

Answer 2

sin(2n)

Answer 3

hmm, any ideas on how to test it?

Answer 4

sin(2n) 1-sqrt n
------ -------
1+sqrt n 1-sqrt n

sin(2n) sqrt n sin(2n)
----- - ------------
1- n 1- n

Answer 5

couldn't you just divide everything by 1+sqrt n? I did -1 is less than equal to sin 2n is less than equal to 1 since sin function oscillates from -1 to 1. Then I ran the limit as n approaches infinity.

Answer 6

let u = 1-n, n = 1-u


sin(2-2u) sqrt n sin(2-2u)
-------- - ------------
u u


sin(2) cos(2u) + cos(2) sin(2u) sqrt n (sin(2) cos(2u) + cos(2) sin(2u))
------------------------- - -------------------------------
u u

Answer 7

youd want to use a conjugate to get rid of the sqrt n part underneath

Answer 8

oh ok that makes sense.

Answer 9

sin(2) cos(2u) + cos(2) sin(2u)
-------------------------
u


2sin(2) cos(2u) + 2cos(2) sin(2u)
-------------------------
2u


cos(2u) sin(2u)
2sin(2) ------- + 2cos(2) --------
2u 2u


---------------------------------------------
the tricky part may be this one if my idea holds

sqrt (1-u) (sin(2) cos(2u) + cos(2) sin(2u))
- -----------------------------------
u

Answer 10

and i see i used a sin(a+b) identity instead of the sin(a-b) so some prudence might be warranted

Answer 11

ok I see what you did. thank you so much!