Derive the Laurent series of $\frac{1}{\sqrt{s^2-a^2}}$ So, I was trying to derive the Laplace Tranform of the Modified Bessel Function Io(at) using its summation formula. I arrived at the answer $$\frac{1}{s}+\frac{a^2}{2s^3}+\frac{3a^4}{8s^5}+\frac{5a^6}{16s^2}+\frac{35a^8}{128s^9}+\frac{63a^{10}}{256s^{11}}+\cdots$$
It wasn't the answer yet so I decided to enter the Laplace Transform of the Modified Bessel function which was $\frac{1}{\sqrt{s^2-a^2}}$ to Wolfram Alpha. According to Wolfram Alpha, its Laurent series  $\frac{1}{\sqrt{s^2-a^2}}$ is equal to the answer I had. 
 $$\frac{1}{s}+\frac{a^2}{2s^3}+\frac{3a^4}{8s^5}+\frac{5a^6}{16s^2}+\frac{35a^8}{128s^9}+\frac{63a^{10}}{256s^{11}}+\cdots$$
I don't know anything about Laurent series. Help
https://www.wolframalpha.com/input/?i=1%2Fsqrt(s%5E2-a%5E2)

Laurent series are quite "similar" to Taylor series. 
Considering that you want the Laurent series of $\frac{1}{\sqrt{s^2-a^2}}$, consider that $$\frac{1}{\sqrt{s^2-a^2}}=\frac 1 s \frac{s}{\sqrt{s^2-a^2}}=\frac 1 s\frac 1{\sqrt{1-\frac{a^2}{s^2}}}$$ Now, by Taylor series or generalized binomial theorem $$\frac 1{\sqrt{1-x^2}}=1+\frac{x^2}{2}+\frac{3 x^4}{8}+\frac{5 x^6}{16}+\frac{35
   x^8}{128}+O\left(x^9\right)$$ Replace $x$ by $\frac a s$ to get $$\frac 1 s\frac 1{\sqrt{1-\frac{a^2}{s^2}}}=\frac 1 s\left(1+\frac{a^2}{2 s^2}+\frac{3 a^4}{8 s^4}+\frac{5 a^6}{16 s^6}+\frac{35 a^8}{128
   s^8}+O\left(\frac{1}{s^9}\right) \right)$$ $$\frac{1}{\sqrt{s^2-a^2}}=\frac{1}{s}+\frac{a^2}{2 s^3}+\frac{3 a^4}{8 s^5}+\frac{5 a^6}{16 s^7}+\frac{35
   a^8}{128 s^9}+O\left(\frac{1}{s^{10}}\right)$$
