This is what I have tried, but I don't know what to do next, so I need help
: $ P(x)=\sum \limits_{k=1}^\infty \arctan\left(\frac{x-1}{(n+x+1)\sqrt{n+1}+(n+2)\sqrt{n+x}}\right). $
$ P(1)=\arctan(\frac{x-1}{(1+x+1)\sqrt{1+1}+(1+2)\sqrt{1+x}})$ $=\arctan(\frac{x-1}{(2+x)\sqrt{2}+3\sqrt{1+x}})$
$ P(2)=\arctan(\frac{x-1}{(2+x+1)\sqrt{2+1}+(2+2)\sqrt{2+x}})$ $=\arctan(\frac{x-1}{(3+x)\sqrt{3}+4\sqrt{2+x}})$
$ P(3)=\arctan(\frac{x-1}{(3+x+1)\sqrt{3+1}+(3+2)\sqrt{3+x}})$ $=\arctan(\frac{x-1}{(4+x)\sqrt{4}+5\sqrt{3+x}})$
$ P(4)=\arctan(\frac{x-1}{(4+x+1)\sqrt{4+1}+(4+2)\sqrt{4+x}})$ $=\arctan(\frac{x-1}{(5+x)\sqrt{5}+6\sqrt{4+x}})$
$ P(5)=\arctan(\frac{x-1}{(5+x+1)\sqrt{5+1}+(5+2)\sqrt{5+x}})$ $=\arctan(\frac{x-1}{(6+x)\sqrt{6}+7\sqrt{5+x}})$
I tried to calculate it for a few numbers to see if there will be any order to conclude something. Only that I conclude is that the series is diverging.
I'll assume $x$ is an positive integer. Define
$$\begin{cases} a_n &= \frac{x-1}{(n+x+1)\sqrt{n+1}+(n+2)\sqrt{n+x}}\\ u_n &= \sqrt{n+x}\\ v_n &= \sqrt{n+1} \end{cases}$$ We have $$a_n = \frac{u_n^2-v_n^2}{(u_n^2+1)v_n + (v_n^2+1)u_n} = \frac{u_n-v_n}{1 + u_nv_n} $$ This leads to
$$\tan^{-1}a_n = \tan^{-1}\left(\frac{u_n-v_n}{1+u_n v_n}\right) = \left(\tan^{-1}u_n - \tan^{-1}v_n\right) + \pi N $$ for some integer $N$ to be determined. By throwing in some explicit numbers, it is not hard to see $N = 0$ in this case. As a result, we have
$$\begin{align} P(x) = \sum_{n=1}^\infty \tan^{-1} a_n &= \sum_{n=1}^\infty \left(\tan^{-1}\sqrt{n+x} - \tan^{-1}\sqrt{n+1}\right)\\ &= \sum_{n=1}^\infty \left( \tan^{-1}\frac{1}{\sqrt{n+1}} - \tan^{-1}\frac{1}{\sqrt{n+x}}\right)\\ &= \sum_{n=1}^{x-1} \tan^{-1}\frac{1}{\sqrt{n+1}} \end{align} $$