Putting the value of the first term, we can see that the series goes like $$1/2, 5/4, 17/16,...$$ I am unable to calculate the general term, and so not sum of the series. Please help me to find the general term or directly the infinite sum if possible.
Let $b_n=a_n-1$ so that $b_0=-\frac 12$ and $b_{n}=b_{n-1}^2$. Consequently, $b_n=b_0^{2^n}$ hence $$a_n=1+(-2)^{-2^n}$$ Let $x=\frac 12$, then \begin{align} (1+x)\prod_{n=0}^Na_n &=(1+x)\prod_{n=0}^N(1+(-x)^{2^n})\\ &=(1+x)(1-x)(1+x^2)\cdots(1+x^{2^N})\\ &=(1-x^2)(1+x^2)\cdots(1+x^{2^N})\\ &=(1-x^4)(1+x^4)\cdots(1+x^{2^N})\\ &=(1-x^8)(1+x^8)\cdots(1+x^{2^N})\\ &=\cdots\\ &=(1-x^{2^N})(1+x^{2^N})\\ &=1-x^{2^{N+1}}\\ &\xrightarrow{N\to\infty}1 \end{align} so that $$\prod_{n=0}^\infty a_n=\frac 23$$