Prove that $\cosh^{-1}(1+x)=\sqrt{2x}(1-\frac{1}{12}x+\frac{3}{160}x^2-\frac{5}{896}x^3+....)$ 
How can we prove the series expansion of
  $$\cosh^{-1}(1+x)=\sqrt{2x}\left(1-\frac{1}{12}x+\frac{3}{160}x^2-\frac{5}{896}x^3+...\right)$$


I know the formula for $\cosh^{-1}(x)=\ln(x+\sqrt{x^2-1})$ so, $$\cosh^{-1}(1+x)=\ln(1+x+\sqrt{x^2+2x}).$$ I tried to apply Maclaurin  series but i could not find the $f(0),f'(0),f''(0),f'''(0)$
Is there any other method available to prove this series expansion like Laurent series etc.
Please help me.Thanks.
 A: I think of $x$ as small.  $\alpha,\beta,\gamma$ are small as well, and depend on $x$ in a way I don't know yet.  $\alpha$ is the biggest term; $\beta$ is next-biggest, then $\gamma$ and so on.  I bring them in as needed.
$$\cosh(\alpha+\beta+\gamma+...)=1+x$$
Leading order: $$1+\frac{\alpha^2}2=1+x+o(x)\text{ , so }\alpha=\sqrt{2x}$$
Next order: $$1+\frac{\alpha^2}2+\alpha\beta+\frac{\alpha^4}{24}=1+x+o(x^2)\text{, so }\beta=-\alpha^3/24=-\sqrt{2x}\frac x{12}$$
A: The trick is to recognize the $1/\sqrt x$ singularity in the derivative of the function of interest and transform the series into a series in $\sqrt x$.  
To proceed, we substitute $z=\sqrt{x}$.  Then, we have
$$\begin{align}
\cosh^{-1}(1+x)&=\cosh^{-1}(1+z^2)\\\\
&=\log\left((z^2+1)+z\sqrt{z^2+2}\right)\tag 1
\end{align}$$
Now, expanding $f(z)=\log\left((z^2+1)+z\sqrt{z^2+2}\right)$ in a series around $z=0$ reveals
$$\begin{align}
f'(z)&=\frac{1}{(z^2+1)+z\sqrt{z^2+2}}\left(2z+\sqrt{z^2+2}+\frac{z^2}{\sqrt{z^2+2}}\right)\\\\
&=\frac{2}{\sqrt{z^2+2}}
\end{align}$$
Then, continuing to differentiate, we find that
$$\begin{align}
f^{(2)}(z)&=-2z(z^2+2)^{3/2}\\\\
f^{(3)}(z)&=4(z^2-1)(z^2+2)^{-5/2}\\\\
f^{(4)}(z)&=-12z(z^2-3)(z^2+2)^{-7/2}\\\\
f^{(5)}(z)&=24(2z^4-12z^3+3)(z^2+2)^{-9/2}
\end{align}$$
Evaluating these at $z=0$ reveals 
$$\begin{align}
f(z)&=f(0)+f'(0)z+\frac12f^{(2)}(0)z^2+\frac16f^{(3)}(0)z^3+\frac1{24}f^{(4)}(0)z^4+\frac1{120}f^{(5)}(0)z^5+O(z^7)\\\\
&=\sqrt 2\,z-\frac{\sqrt{2}}{12}z^3+\frac{3\sqrt 2}{160}z^5+O(z^7)\\\\
&=\sqrt{2x}\left(1-\frac1{12}x+\frac{3}{160}x^2+O(x^3)\right)
\end{align}$$
A: I you just want to find the first few terms, Then consider getting a power series for the function
$$g(x) = \frac{\cosh^{-1}(1+x)}{\sqrt{x}}=\frac{\ln (1+x+\sqrt{x^2+2x})}{\sqrt{x}}$$ by taking derivatives and evaluating at $x=0$. Then you can multiply the power series by $\sqrt{x}$.
There is some circulation in this argument in the sense that the choice of $\sqrt x$ came from knowing what the answer is. It is not uncommon to chose an arbitrary power of $x$ at first, and see what works.
The problem with this approach is that you have to use L'Hopital's rule every time you evaluate a derivative, by taking a limit at zero.
The smarter way is to use asymptotic expansions, which is pretty much what @Michael did in his answer.
