# Thread: series proof

1. ## series proof

hello I am given prove that

$\displaystyle \frac{sin\theta}{2} + \frac{sin2\theta}{2^2} + \frac{sin3\theta}{2^3} +...........= \frac{2sin\theta}{5-4cos\theta}$

what I have done
this is a Geometric progression therefore sum to infinity = $\displaystyle \frac{a}{1-r}$

a= $\displaystyle \frac{sin\theta}{2}$

r = $\displaystyle \frac{sin2\theta}{2^2} / \frac{sin\theta}{2}$
r = $\displaystyle \frac{sin2\theta}{2sin\theta}$

substituting back into the equation I get

sum to infinity = $\displaystyle \frac{2sin^2\theta}{2-2cos\theta}$

can someone please show me where I am going wrong, thanks.

2. ## Re: series proof

Originally Posted by sigma1
hello I am given prove that

$\displaystyle \frac{sin\theta}{2} + \frac{sin2\theta}{2^2} + \frac{sin3\theta}{2^3} +...........= \frac{2sin\theta}{5-4cos\theta}$

what I have done
this is a Geometric progression therefore sum to infinity = $\displaystyle \frac{a}{1-r}$

a= $\displaystyle \frac{sin\theta}{2}$

r = $\displaystyle \frac{sin2\theta}{2^2} / \frac{sin\theta}{2}$
r = $\displaystyle \frac{sin2\theta}{2sin\theta}$

substituting back into the equation I get

sum to infinity = $\displaystyle \frac{2sin^2\theta}{2-2cos\theta}$

can someone please show me where I am going wrong, thanks.
It's not geometric, simplify the 2nd and 3rd terms using sin(A + B) = sinAcosB + cosAsinB to compare with 1st term sine function

$\displaystyle \frac{sin \theta}{2} \ + \ \frac{sin \theta cos \theta}{2} \ + \ \frac{sin \theta (4cos^2 \theta -1)}{8} \ + \ ...$

You don't have a common ratio here

3. ## Re: series proof

Originally Posted by agentmulder
It's not geometric, simplify the 2nd and 3rd terms using sin(A + B) = sinAcosB + cosAsinB to compare with 1st term sine function

$\displaystyle \frac{sin \theta}{2} \ + \ \frac{sin \theta cos \theta}{2} \ + \ \frac{sin \theta (4cos^2 \theta -1)}{8} \ + \ ...$

You don't have a common ratio here

thank you for pointing that out. it would be a great help if you could show me how you would attempt to further solve this.

4. ## Re: series proof

Your problem requires cleverness i do not have at this time so i am looking forward to a solution posted by another MHF member. The series is interesting.

WIA is also stumped.

series&#40;sin&#40;nx&#41;&#41;&#47;&#40;2&#94;n&# 41; , n &#61; 1 to infinity&#41; - Wolfram|Alpha

5. ## Re: series proof

Originally Posted by sigma1
hello I am given prove that

$\displaystyle \frac{sin\theta}{2} + \frac{sin2\theta}{2^2} + \frac{sin3\theta}{2^3} +...........= \frac{2sin\theta}{5-4cos\theta}$

what I have done
this is a Geometric progression therefore sum to infinity = $\displaystyle \frac{a}{1-r}$

a= $\displaystyle \frac{sin\theta}{2}$

r = $\displaystyle \frac{sin2\theta}{2^2} / \frac{sin\theta}{2}$
r = $\displaystyle \frac{sin2\theta}{2sin\theta}$

substituting back into the equation I get

sum to infinity = $\displaystyle \frac{2sin^2\theta}{2-2cos\theta}$

can someone please show me where I am going wrong, thanks.
You could probably use induction.

6. ## Re: series proof

I'll replace $\displaystyle \theta$ with $\displaystyle x$ for convenience.

Using the identity $\displaystyle \sin(x)=i\frac{e^{-ix}-e^{ix}}{2}$, your series may be rewritten as

$\displaystyle \displaystyle{\sum_{n=1}^{\infty}\frac{i}{2^{n+1}} e^{-inx}}-\displaystyle{\sum_{n=1}^{\infty}\frac{i}{2^{n+1}} e^{inx}}$.

Individually, these two sums are convergent geometric series (do you know why?). Calculate the sum for each term and add them up. I have verified that you do indeed get the desired result (after some algebraic manipulations), but I have no desire to type it out. You should be able to work it out yourself.

You may need to justify why you are able to break the sum into two infinite series, but this is a straightforward application of limit theorems since both your series converge.