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)) + ... + 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) - 1/3 k^(-3/2) + 1/5 k^(-5/2) + ...

So summing over 1 to n, we should get

f(n) = (1^(-1/2) + 2^(-1/2) + ... + n^(-1/2))

• 1/3 (1^(-3/2) + 2^(-3/2) + ... + n^(-3/2))

• 1/5 (1^(-5/2) + 2^(-5/2) + ... + n^(-5/2))

• ...

Now, 1^(-1/2) + 2^(-1/2) + ... + 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. Zeta is the Riemann zeta function. And 1^(-j/2) + 2^(-j/2) + ... + 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) - 1/3 Zeta(3/2) + 1/5 Zeta(5/2) - ...) + 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 φ, 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.

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.