How to approximate sqrt(A+B)

Assume that [itex]A,B\in\mathbb{R}^{N\times N}[/itex] are some matrices which do not commute. Is there any way to approximate the matrix



[tex]

\sqrt{A + B}

[/tex]



under the assumption that [itex]B[/itex] is smaller than [itex]A[/itex]? Could it be possible to write the square root precisely with some infinite series under some conditions?



Here's a related problem, which I know already. Suppose we were interested in the matrix



[tex]

(A + B)^{-1}

[/tex]



We know for complex numbers [itex]|z|<1[/itex] the series



[tex]

\frac{1}{1 + z} = 1 - z + \sum_{n=2}^{\infty}(-1)^n z^n

[/tex]



The same series will work for bounded operators with norm less than one, so by using the formula [itex](XY)^{-1}=Y^{-1}X^{-1}[/itex] we get



[tex]

(A + B)^{-1} = ((\textrm{id} + BA^{-1})A)^{-1} = A^{-1}(\textrm{id} + BA^{-1})^{-1}

[/tex]

[tex]

= A^{-1}\Big(\textrm{id} - BA^{-1}+ BA^{-1}BA^{-1} + \sum_{n=3}^{\infty}(-1)^n(BA^{-1})^n\Big)

[/tex]

[tex]

= A^{-1} - A^{-1}BA^{-1}+ A^{-1}BA^{-1}BA^{-1} - \cdots

[/tex]



We also have Taylor series for square root:



[tex]

\sqrt{1 + z} = 1+ \frac{1}{2}z- \frac{1}{4}z^2 + \frac{3}{8}z^3

+ \sum_{n=4}^{\infty}\frac{(-1)^{n+1}3\cdot 5\cdot\ldots\cdot (2n-3)}{2^n}z^n

[/tex]



Could it be, that this series could be used get something for the [itex]\sqrt{A+B}[/itex]? Actually the previous trick doesn't work now. The problem is that [itex]\sqrt{XY}[/itex] is not related to [itex]\sqrt{X}\sqrt{Y}[/itex] or [itex]\sqrt{Y}\sqrt{X}[/itex] in any obvious way. So I was left without further ideas here.