Calculate the length of the curve. $L_1, \ L_2,...,L_{12}$ are twelve lines going through origo with $30^{\text{o}}$ apart as shown in the figure. The length of $OP_1$ is equal to $1$. One can, as shown in the figure construct an infinite spiral by drawing normals from the next line to the previous. If this spiral is to go around origo for an infinite number of times, what is the total length of this spiral curve?


I think I solved this problem, but my book has no answer page, so I need someone to check my work. This is what I did:
I noticed that
\begin{array}{lcl}
P_1P_2 & = & OP_1\sin{30} \\
P_2P_3 & = & OP_2\sin{30} =OP_1\cos{30}\sin{30} \\
P_3P_4 & = & OP_2\cos{30}\sin{30}=OP_1\cos^2{30}\sin{30} \\
 & \vdots & \\
P_{n+1}P_{n+2} & = & \cos^{n}{30}\sin{30}.
\end{array}
So, the total length of the curve is given by the geometric sum $$\sum_{k=1}^{\infty}\cos^k{30}\sin{30}=\lim_{n\rightarrow\infty}\frac{1}{2}\sum_{k=1}^{n}\left(\frac{\sqrt{3}}{2}\right)^k=\lim_{n\rightarrow\infty}\frac{1}{2}\frac{\left(\frac{\sqrt{3}}{2}\right)^{n+1}-\left(\frac{\sqrt{3}}{2}\right)}{1-\frac{\sqrt{3}}{2}}= \frac{\sqrt{3}}{4-2\sqrt{3}}\approx3.232.$$
Questions:


*

*Is the above correct?

*Why do I get a different answer when I instead use $P_{n}P_{n+1}  =  \cos^{n-1}{30}\sin{30}?$ I then get


$$\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{\sqrt{3}}{2}\right)^k\left(\frac{4}{\sqrt{3}}\right)=\frac{2}{2-\sqrt{3}}\approx7.465.$$
Note: I don't want an alternative fance show-off-solution, I just want to know how I can improve my method.
 A: *

*Your original response is incorrect, as you should be taking the sum from $k=0$ to infinity. An easy way to convince yourself of this is that your sum must include the term $P_1P_2=\sin(30)$, so we must include the $k=0$ term of $\cos(30)^k \sin(30)$. Doing this, we evaluate your limit again as 
$$\sum_{\color{red}{k=0}}^{\infty}\cos^k(30)\sin(30)=\lim_{n\rightarrow\infty}\frac{1}{2}\sum_{\color{red}{k=0}}^{n}\left(\frac{\sqrt{3}}{2}\right)^k=\lim_{n\rightarrow\infty}\frac{1}{2}\frac{\color{red}{1-\left(\frac{\sqrt{3}}{2}\right)^{n+1}}}{1-\frac{\sqrt{3}}{2}}= \frac{1}{2-\sqrt3}=2+\sqrt3\approx3.732$$
Note that here you messed up the sign for the geometric sum, which I highlighted in red. You made another sign error which made up for it, but it is better to have the correct expression at each step.

*Now when you used $P_nP_{n+1}$ as $\cos^{n-1}(30)\sin(30)$, we do start from $k=1$ in our summation as again, we must include the term $P_1P_2=\sin(30)$. Your problem here is that in your summation you should have used $\frac{2}{\sqrt{3}}$ instead of $\frac{4}{\sqrt{3}}$, but as you do not give an explanation for why you chose $4$, I cannot explain why you are incorrect.
A: As we reduce $30^{\circ}$ interval to zero, length of curve, the log spiral 
$$ r= OP_1 e ^{-\theta/\sqrt3} $$
is bounded to a finite limit evaluated by definite integral
$$ OP_1 ( 4/\sqrt3)$$
as we go to center from $1$ to $n.$
