Question: Let $\sum\_{k=0}^\infty a\_k x^k$ be the power series representaion of a real valued function $f(x)$ in $(-1,1).$ Is there any necessary and sufficient condition on the coefficients so that $f(x)\geq 0$ for all $x\in [0,1).$ An obvious sufficient condition is $a\_n\geq 0$ for all $n.$

Any help or reference will be appreciated.

Answer: 
Even the necessary and sufficient condition that a polynomial expression is positive of a given interval becomes very complicated as the degree of the polynomial increases. However, there are sufficient condition, for instance the SOS (sums of squares) method. In some cases, the conditions are necessary and sufficient.

As for series, that becomes even more difficult I guess. 

Here is an example of a positive series on $\mathbb{R}$

$$1 - \cos x = \sum\_{n\ge 1}(-1)^{n-1}\frac{x^{2n}}{(2n)!}$$

If you didn't know anything about trigonometric functions, proving this inequality can be done by showing that the series is a square of another series (in this case $\sqrt{2} \sin \frac{x}{2})$ that you are lucky to guess.
(A bit too long for a comment:) In this general context it seems too difficult. The first non-zero coeff obviously must have an even index $a\_{2j}$, so factoring by $a\_{2j} x^{2j}$ it suffices to look at the factor so you may reduce to the case $a\_0=1$. 

Now, given $a\_0=1$ and any values of $a\_1,...,a\_{2d-1}$ let $p(x)=\sum\_{k=0}^{2d-1} a\_k x^k$. If this is not non-negative on $(-1,1)$ you may set
$$ a\_{2d}=- \inf\_{0<|x|<1} x^{-2d} P(x) \;.$$
The inf is finite since $P>0$ on an interval $(-\delta,\delta)$ for some $\delta>0$.
Then $p(x) + a\_{2d} x^{2d}$ is non-negative on $(-1,1)$. So you might want to put further constraints, e.g. the coeffs being uniformly bounded? or perhaps better: $f(z)$ being uniformly bounded on the unit disk?