Relativistic Bohr Atom and MacLaurin Series

[atomicpedals]

Homework Statement

By expanding a MacLaurin Series show that
E_{n}=\epsilon_{n} - \mu c^{2} = - \frac{w_{0}Z^{2}}{n^{2}}[1+\frac{\alpha^{2} Z^{2}}{n}(\frac{1}{k}-\frac{3}{4n})]

Homework Equations

Through a lengthy derivation I arrived at
\epsilon_{n}=\frac{\mu c^2}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{n_{r}+\sqrt{l^{2}-Z^{2}\alpha^{2}}}}}
I should add that k is what the text is using for the azimuthal quantum number, I used l in my derivation out of habit.

The Attempt at a Solution

I've got no ideas where to go with this thing. I should take advantage of identites
\sqrt{1-x}=1-\frac{x}{2}-\frac{x^{2}}{8}+...
\frac{1}{1+x}=1+...
Do I need to make some aggressive substitutions?

 

[atomicpedals]

I think I can make the following justifiable substitution
\epsilon_{n}=\frac{\mu c^2}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{n'}}}
Where n' = n_{r} + \sqrt{l^{2}-Z^{2} \alpha^{2}}. But this still doesn't get me any closer to arriving at E_{n}=\epsilon_{n} - \mu c^{2} = - \frac{w_{0}Z^{2}}{n^{2}}[1+\frac{\alpha^{2} Z^{2}}{n}(\frac{1}{k}-\frac{3}{4n})] through a Maclaurin series.

 

[atomicpedals]

Ok, I may be closer... using the substitution above to put E_{n} = \epsilon_{n} - \mu c^{2} in terms of n' I can calculate a Maclaurin series as follows
- \mu c^{2} + \frac{\mu c^{2} \sqrt{n'}}{\alpha Z} - \frac{n^{3/2} (\mu c^{2} \alpha Z)}{2(\alpha^{4} Z^{4})} + \frac{n^{5/2} (3 \mu c^{2} \alpha Z)}{8 \alpha^{6} Z^{6}} - ...
Am I on the right track to the final E_{n}?