Prove that a sequence is Cauchy in an inner product space.

Question

I need to show that the space $H=B(\mathbb{R},\mathbb{C})$ of complex bounded real-valued functions equipped with the inner product $$\langle f,g\rangle=\int_{\mathbb{R}}\frac{f(x)\overline{g(x)}}{1+x^2}dx$$ is not a Hilbert space. To show this the hint was to use the sequence of functions $f_n(x):=\min \{x^{-\frac{1}{4}},n^{\frac{1}{4}}\}$ for all $x\geq 0$ and $0$ for $x<0$. The limit $f(x)=x^{-\frac{1}{4}}\chi_{(0,+\infty)}(x)\notin H$, I need to prove that $\Vert f_m-f_n\Vert\to 0$ as $m,n\to +\infty$. Let for example $m\geq n$. We have $$\Vert f_m-f_n\Vert^2=\langle f_m-f_n,f_m-f_n \rangle=\int _0^\frac{1}{m}\frac{(m^\frac{1}{4}-n^\frac{1}{4})^2}{1+x^2}dx+\int_\frac{1}{m}^\frac{1}{n}\frac{(x^\frac{1}{4}-n^\frac{1}{4})^2}{1+x^2}dx$$ Because the difference function $f_m-f_n$ is equal to $m^\frac{1}{4}-n^\frac{1}{4}$ if $0\leq x\leq\frac{1}{m}$, $x^\frac{1}{4}-n^\frac{1}{4}$ if $\frac{1}{m}\leq x\leq\frac{1}{n}$ and $0$ otherwise. For the first term on the right one can write the explicit value of the integral easily and, having supposed $m\geq n$, we have $$\int _0^\frac{1}{m}\frac{(m^\frac{1}{4}-n^\frac{1}{4})^2}{1+x^2}dx= (m^\frac{1}{4}-n^\frac{1}{4})^2\arctan\frac{1}{m}\leq m^\frac{1}{2}\arctan\frac {1}{m}\sim m^{-\frac{1}{2}}$$ as $m\to+\infty$, so it converges. I don't know how to deal with the second term, I tried something but every time the result seems to be dependent by the fact that $m$ goes to infinity faster than $n$. Does somebody have any suggestion?