Solution of Equation $ 1 + x + \frac{1\cdot3}{2!}x^2 + \frac{1\cdot3\cdot5}{3!}x^3 + \ldots = \sqrt{2} $ I have an equation whose left side is a infinite series . I can solve the equation if I am able to find a close form of the series . The equation is as follows :  
$$ 1 + x + \frac{1\cdot3}{2!}x^2 + \frac{1\cdot3\cdot5}{3!}x^3 + \ldots = \sqrt{2} $$  
I have tried to find the sum of the series by trying to apply binomial theorm of negative exponent . But I cant solve this problem . Can you help me to find the sum of series of the left hand side of the above equation ? 

Compare the series with $1+na+\frac{n(n-1)a^2}{2!}\cdots$ we get
$na=x$ or $$n^2a^2=x^2\dots(1)$$
also $$\frac{n(n-1)a^2}{2!}=\frac{1.3x^2}{2!}\cdots(2)$$
Divide $(1)$ and $(2)$ and solve for $n$, substitute in first and solve for $a$
you will get the sum to be $(1-2x)^{\frac{-1}2}$
