spiral of Theodorus A long time ago when I was in college I read about making a spiral out of right triangles with sides 1 and $\sqrt{N}$. (A google search seems to indicate that this is called the Spiral of Theodorus.)

I spent a long time trying to prove that the series of points approximated a spiral $R = K\theta + \varphi$, by trying to show the limit of the difference $\varphi = \sqrt{N+1} - K \sum_{i=1}^N \arctan(1/\sqrt{i})$ existed for some $K$. I think I managed to do it but it was confusing and can't find my papers. (and I'm still an amateur mathematician!)
Is this a known problem, and is there a closed-form solution to $K$ and $\varphi$?
 A: Here's a sketch of a proof that the constant you want exists, and how to find it.
Let 
$$
f(n) = \arctan(1) + \arctan(1/\sqrt{2}) + \arctan(1/\sqrt{3}) + \ldots + \arctan(1/\sqrt{n}).
$$
You want to show that $f(n) = \sqrt{n} + C + o(1)$ for some constant $C$.  (If you're not familiar with the $o$-notation, think of $o(1)$ as representing some function which goes to $0$ as $n$ goes to infinity.)  
Then take the power series expansion of $\arctan(1/\sqrt{k})$; this is
$$
(*) ~~~~~~~k^{-1/2} - \frac{1}{3} k^{-3/2} + \frac{1}{5} k^{-5/2} + \ldots
$$
So summing over $1$ to $n$, we should get
\begin{align*}
f(n)  = & (1^{-1/2} + 2^{-1/2} + ... + n^{-1/2}) \\\
  - \, \frac{1}{3} &(1^{-3/2} + 2^{-3/2} + ... + n^{-3/2}) \\\
 + \, \frac{1}{5}& (1^{-5/2} + 2^{-5/2} + ... + n^{-5/2}) - \ldots 
\end{align*}
Now, $1^{-1/2} + 2^{-1/2} + \ldots + n^{-1/2}$ has the asymptotic form
$$
2 \sqrt{n} + \zeta(1/2) + O(n^{-1/2}) 
$$
where I cheated a bit and asked Maple, and $\zeta$ is the Riemann zeta function.  And $1^{-j/2} + 2^{-j/2} + \ldots + n^{-j/2}$ has the asymptotic form
$$
\zeta(j/2) - O(n^{-j/2 + 1})
$$
where, if you're not familiar with the $O$-notation, $O(n^{-j/2+1})$ should be thought of as a function that goes to zero at least as fast as $n^{-j/2 + 1}$ as n goes to infinity.  So, assuming that we can rearrange series however we like,
$$
f(n) = 2 \sqrt{n} + (\zeta(1/2) - \frac{1}{3} \zeta(3/2) + \frac{1}{5} \zeta(5/2) - \ldots) + o(1).
$$
Since $\zeta(s)$ is very close to $1$ when $s$ is a large real number, that alternating series should converge.  Again cheating and using Maple, I claim it converges to about $−2.157782997$.  This is the constant you call $\varphi$, and what you called $K$ is equal to $2$.  (An easier way to see that your $K$ is $2$ is to note that $\arctan(1/\sqrt{n})$ is about $1/\sqrt{n}$, and approximate the sum by an integral.
