# series proof

• Apr 9th 2013, 07:55 PM
sigma1
series proof
hello I am given prove that

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

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

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

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

substituting back into the equation I get

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

can someone please show me where I am going wrong, thanks.
• Apr 9th 2013, 11:12 PM
agentmulder
Re: series proof
Quote:

Originally Posted by sigma1
hello I am given prove that

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

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

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

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

substituting back into the equation I get

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

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

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

You don't have a common ratio here

:)
• Apr 9th 2013, 11:25 PM
sigma1
Re: series proof
Quote:

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

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

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.
• Apr 10th 2013, 12:21 AM
agentmulder
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
• Apr 10th 2013, 12:23 AM
Prove It
Re: series proof
Quote:

Originally Posted by sigma1
hello I am given prove that

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

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

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

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

substituting back into the equation I get

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

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

You could probably use induction.
• Apr 10th 2013, 07:18 AM
Gusbob
Re: series proof
I'll replace $\theta$ with $x$ for convenience.

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

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

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.