What Values of x Allow Convergence in a Geometric Sequence of Sine Functions?

[Matejxx1]

Homework Statement

given a geometric sequence sin(x),sin(2x), . . .
c) find for which values of x∈(0,π) this sequence converges and calculate its limit

Homework Equations

|q|<1 or -1<q<1

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

 

[PeroK]

What makes you think that is s geometric sequence?

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

 

[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

 

[HallsofIvy]

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).

 

[PeroK]

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

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

 

[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

 

[Matejxx1]

And the sequence is sin(x), 2 cos(x)sin(x), 4cos^2(x)sin(x), 8 cos^3(x)sin(x), ... ?

yes that's the sequence.

converges to a/(1- r). Here you are saying that a= sin(x) and r= 2cos(x).

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
Could someone explain the equal sign in the first inequality

 

[PeroK]

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

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

 

[Matejxx1]

yes basically

 

[PeroK]

yes basically

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

 

[Matejxx1]

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

 

[PeroK]

Many sequences converge when the corresponding series does not.

What about #q =1##?

 

[Matejxx1]

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

 

[Ray Vickson]

Homework Statement

given a geometric sequence sin(x),sin(2x), . . .
c) find for which values of x∈(0,π) this sequence converges and calculate its limit

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), ...?

 

[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, . . .

 

[Matejxx1]

Many sequences converge when the corresponding series does not.

What about #q =1##?

i think I get it now
the sequence
a_n=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 doesn't exist here either
am I correct about this?

 

[PeroK]

i think I get it now
the sequence
a_n=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 doesn't exist here either
am I correct about this?

Yes.