Find the limit of a sequence

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1. Jan 21, 2016

Matejxx1

1. The problem statement, all variables and given/known data
given a geometric sequence sin(x),sin(2x), . . .
c) find for which values of x∈(0,π) this sequence converges and calculate its limit

2. Relevant equations
|q|<1 or -1<q<1

3. The attempt at a solution
Ok so in part a) and b) i calculated the quotient and found out that
q=sin2x/sinx
q=(2*cos(x)*sin(x))/(sin(x))
q=2*cos(x)
so know i tried to figure out for which values x will converge
|q|<1
-1<q<1
-1<2*cos(x)<1 /:2
-1/2<cos(x)<1/2
2π/3>x>π/3
which means that the sequence will converge when the angle is anywhere between 120° and 60°
However i'm unsure now how to continue to find the limit because I've never seen an example where the value x isin't exactly defined
could someone show me how this is done or give me some kind of a hit or check if I made a mistake somewhere inbetween
Thanks

Last edited: Jan 21, 2016
2. Jan 21, 2016

PeroK

What makes you think that is s geometric sequence?

Also, you use both the terms series and sequence. Which is it?

3. Jan 21, 2016

Matejxx1

I'm really sorry I made a mistake while translating the text into english.
Also it is written in the problem that this is a geometric sequence

4. Jan 21, 2016

HallsofIvy

Staff Emeritus
And the sequence is sin(x), 2 cos(x)sin(x), 4cos^2(x)sin(x), 8 cos^3(x)sin(x), ... ?
Yes, the "geometric sequence a, ar, ar^2, ar^3, ... converges for -1< r< 1 and converges to a/(1- r). Here you are saying that a= sin(x) and r= 2cos(x).

5. Jan 21, 2016

PeroK

Perhaps there is a problem in translation, but clearly $sin(nx)$ is not in general a geometric sequence. Take $x = \pi/2$ for example.

6. Jan 21, 2016

Matejxx1

Perhaps it will be better if I write the whole problem out not just the last step I'll try to do that right now

7. Jan 21, 2016

Matejxx1

yes that's the sequence.
but woudn't that just give me the sum of all elements of the sequence?
Edit :
I asked my profesor if he could give me the solutions and he didn't have much time but I did get to take a picture if this helps in any way
so the solutions to 5 c)
are π/3 ≤ x < 2*π/3 --------- it is the same that I got but I don't know why there is an equal sign?
x=π/3 ----------- the limit is sqrt(3)/2
π/3 < x < 2*π/3 the limit is 0 <------------------- i get this because if -1< r< 1 the sequence converges to 0

8. Jan 21, 2016

PeroK

So, the question is: For which values of x does the geometric sequence converge:

$sinx, sinx(2cosx), sinx(2cosx)^2 \dots$

9. Jan 21, 2016

Matejxx1

yes basically

10. Jan 21, 2016

PeroK

Why do you think $|q| < 1$?

11. Jan 21, 2016

Matejxx1

Well, I though if the series converges the sequence would converge as well.
Therefore I went ahead and wrote that |q|<1

12. Jan 21, 2016

PeroK

Many sequences converge when the corresponding series does not.

13. Jan 21, 2016

Matejxx1

Hmmm, now that you pointed it out if q=1 the limit of the sequence would simply be sin(x)

14. Jan 21, 2016

Ray Vickson

Is the sequence
sin(x), sin(2x), sin(3x), sin(4x), sin(5x), ...
or is it
sin(x), sin(x), sin(4x), sin(8x), sin(16x), ....?

15. Jan 21, 2016

Matejxx1

the sequence is
sin(x), sin(x)*2cos(x), sin(x)*(2*cos(x))2, sin(x)*(2*cos(x))3, sin(x)*(2*cos(x))4, . . .

16. Jan 21, 2016

Matejxx1

i think I get it now
the sequence
an=sin(x)*(2*cos(x))n-1 will converge when either |2*cos(x)|<1--------- when that is true the sequence will converge to 0
or when 2*cos(x)=1
2*cos(x)>1 this is not possbile because a limit would not exist here
2*cos(x)=-1 because the limit doesnt exist here either