If $A_n$ is a sequence of positive bounded linear operators converging in norm to $A$ on a Hilbert Space, show $\sqrt{A_n}\to\sqrt{A}$ in norm. If $A_n$ is a sequence of positive bounded linear operators converging in norm to $A$ on a Hilbert Space, show $\sqrt{A_n}\to\sqrt{A}$ in norm. I can show that $A$ would be positive and  thus have a square root, but then I'm mostly stuck.
If $A_n=B_n^2$ and $A=B^2$, I have also shown that since $B_n$ is positive it is by definition self adjoint and so
$$\|B_n x\| = \sqrt{\langle B_n x, B_n x \rangle} = \sqrt{\langle A_n x, x\rangle} \to \sqrt{\langle Ax,x\rangle} = \|Bx\|2$$ for all $x$ and so therefore $\|B_n\|\to \|B\|$. However, I am completely stuck on the desired result.
If I knew $B_n$ and $B$ would commute, then I'd use
$$\|A_n^2 - A\| = \|(B_n - B)(B_n + B) \|$$ and play with the inner product, but I don't know this a priori. 
Thanks for your help.
EDIT: For those wondering this question comes from Mathematical Physics I: Functional Analysis by Reed and Simon in chapter 7 question 14.
 A: There is nothing particular about the square root here, as the result holds for any continuous function. 
From $A_n\to A$, we get in particular that $\|A_n\|\to\|A\|$. This guarantees that $\sigma(A)\cup\bigcup_n\sigma(A_n)$ is contained in some interval $[a,b]$. 
Let $f:[a,b]\to\mathbb R$ be continuous. Using the Spectral Theorem, we get that $\|f(X)\|\leq\|f\|_\infty$ for any selfadjoint $X$ with $\sigma(X)\subset[a,b]$. Indeed,
$$
\left|\langle f(X)x,x\rangle\right|=\left|\langle \int_{\sigma(X)}\,f(\lambda)\,dE_X(\lambda)\,x,x\rangle\right|=\left|\int_{\sigma(X)}\,f(\lambda)\,\langle dE_X(\lambda)x,x\rangle
\right|\\ \leq\int_{\sigma(X)}\,|f(\lambda)|\,\langle dE_X(\lambda)x,x\rangle\leq\|f\|_\infty\int_{\sigma(X)}\,1\,\langle dE_X(\lambda)x,x\rangle=\|f\|_\infty\,\langle x,x\rangle.
$$
(the same estimate can be obtained via the Gelfand Transform if one doesn't want to use the Spectral Theorem). 
Now fix $\varepsilon>0$. Let $p$ be a polynomial such that $\|f-p\|_\infty<\varepsilon/3$ (this polynomial exists by Weierstrass Approximation Theorem). As $A_n^k\to A^k$ for all $k$, there exists $n_0$ such that $\|p(A_n)-p(A)\|<\varepsilon/3$ if $n\geq n_0$. 
So, for $n\geq n_0$, 
$$
\|f(A_n)-f(A)\|\leq\|f(A_n)-p(A_n)\|+\|p(A_n)-p(A)\|+\|p(A)-f(A)\|\\ \leq\|f-p\|_\infty+\frac\varepsilon3+\|p-f\|_\infty<\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon.
$$
