Obtain the expression $T=\frac x2+\sum_{n=2}^\infty (-1)^{n-1} \frac{1*3*\ldots*(2n-3)}{n!}*\frac{x^n}{2^n}$ Obtain the expression
$$T=\frac x2+\sum_{n=2}^\infty (-1)^{n-1} \frac{1*3*\ldots*(2n-3)}{n!}*\frac{x^n}{2^n}$$
for one root of the equation
$$T^2+2T-x=0,$$
and show it converges so long as $|x| \lt 1$.  
I have no idea where to start with this, can I have a hint on how to obtain the expression? and how do I find the other root? Can I use the ratio test on the expression to show that it converges?

First notice that 
\begin{align}
T(x) &= \frac{x}{2} - \frac{1}{2} \, \sum_{n=2}^{\infty} (-1)^{n-1} \frac{(-1/2)_{n} \, x^{n}}{n!} \\
&= \frac{x}{2} + \frac{1}{2} \, \left[ (1+x)^{-1/2} - x -2  \right] \\
&= -1 + \sqrt{1+x}
\end{align}
Now the quadratic equation $T^{2} + 2 \, T - x = 0$ leads to
$$T(x) = - 1 \pm \sqrt{1+x}$$
and is of the same form as the series for $T(x)$ given in the problem.
As to the convergence it is not hard to show by any number of standard tests and is left do to.
