Show Cauchy Sequence Convergence with sin(n)/2^n

[sara_87]

Homework Statement

Show that the following sequence converges:

xn= (sin(1)/2) + (sin(2)/2^2) + (sin(3)/2^3) +...+ (sin(n)/2^n)

Homework Equations

The Attempt at a Solution

To show that it converges, i want to show that it is a cauchy sequence (since all cauchy sequences converge).
I know that xn is cauchy if abs(xn-xm)< E for all E>0.
and the above sequence can be written as:
$$\sum(sink/2^k$$

But i don't know how to proceed??
Any help would be very much appreciated.

 

[dirk_mec1]

(since all cauchy sequences converge).

Is that a fact? Check really if that's true...

As for your sequence, your sequence is (monotonically) increasing and bounded thus convergent. Can you show that it is true?

 

[Dick]

Is that a fact? Check really if that's true...

As for your sequence, your sequence is (monotonically) increasing and bounded thus convergent. Can you show that it is true?

It's not monotonic. sin(k) can be negative. And why are you questioning whether cauchy sequences converge?? The question is clearly about the real numbers, and they are complete. All cauchy sequences DO converge. sara_87, to show it's cauchy just notice sin(k)/2^k<=1/2^k. You can bound abs(xn-xm) by the sum of a geometric series.

 

[dirk_mec1]

It's not monotonic. sin(k) can be negative. And why are you questioning whether cauchy sequences converge?? The question is clearly about the real numbers, and they are complete. All cauchy sequences DO converge.

Yes, you're right I made two mistakes.

sara_87, to show it's cauchy just notice sin(k)/2^k<=1/2^k. You can bound abs(xn-xm) by the sum of a geometric series.

Dick uses $$sin(x) \leq 1$$

 

[Dick]

Yep. And I should have written |sin(k)/2^k|<=1/2^k.

 

[sara_87]

ok, so |sin(k)/2^k|<=1/2^k

and the formula for geometric series is:
a(1-r^n)/1-r

i don't know how to apply this to sin(k)/2^k

 

[dirk_mec1]

Look, you want to have $$|a_n - a_m|< \varepsilon$$.

Now you know you can make an upper bound via a geometric series, okay?

 

[sara_87]

Yes, i understand this bit and i know that in this case, we have:
abs(sin(k)/2^k) < 1/2^k

but is this it?? surely not. I still haven't proved that it is cauchy.

 

[Mark44]

Yes, i understand this bit and i know that in this case, we have:
abs(sin(k)/2^k) < 1/2^k

but is this it?? surely not. I still haven't proved that it is cauchy.

The problem as stated asked that you prove that the sequence was convergent. You brought up the part about Cauchy sequences. If you can show that the sequence converges (see Dick's and Dirk's posts), you're done.

 

[sara_87]

Yes, but i 'want' to show that it is cauchy (part of the question). I know that:
abs(xn-xm)< E
and
abs(sin(k)/2^k) < 1/2^k
but how do i relate these two?

 

[Mark44]

Using your definition for x_n at the beginning of this thread, what is x_n - x_m?

 

[sara_87]

(sin(1)/2) + (sin(2)/2^2) + (sin(3)/2^3) +...+ (sin(n)/2^n) - xm

 

[Dick]

Let's assume n>m. So xn-xm=sin(m+1)/2^(m+1)+sin(m+2)/2^(m+2)+...sin(n)/2^n. |xn-xm|<=1/2^(m+1)+...+1/2^n. Apply your geometric series thing to that.

 

[sara_87]

ok, so i get:
abs(xn-xm)<=1/2^m for all m>N ;
but this must mean that 1/2^m is less than E but how do we know that m is greater than E (for this to be true)?

 

[Dick]

You are given e. You PICK an N large enough that 1/2^N<e. That means |xn-xm|<e for all n,m>N, right? That's Cauchy.

 

[sara_87]

but however large N is, the 1/2^N will always be positive.
so, just as an example, what kind of value can N be?

 

[dirk_mec1]

N will depend on your epsilon chosen. In your case you should use the geometric series to find your N.

 

[Dick]

but however large N is, the 1/2^N will always be positive.
so, just as an example, what kind of value can N be?

Pick N so 2^N>1/epsilon.