ca.analysis and odes - spiral of Theodorus

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.

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