i want to prove this identity:
$(1 + \sum\limits_{n=1}^\infty {1/2 \choose n} X^n)^2 = 1+X$
in the formal power series ring Q[[X]]. (so i can't just quote the binomial expansion for the square root)
the only thing i can think of is to calculate the coefficients of each power of x on the left hand side and work them out directly, but it gets very messy and painful and life's too short for it.
anyone care to bash through the calculations or provide insight? (not homework, by the way)
Best Answer
The Chu-Vandermonde Identity gives that $$ \sum_{k=0}^n\binom{1/2}{k}\binom{1/2}{n-k}=\binom{1}{n}\tag{1} $$ Computing the square of the series using $(1)$ and the product formula for the ring structure of the formal power series ring $$ (a_n)_{n\in\mathbb{N}}\times(b_n)_{n\in\mathbb{N}}=\left(\sum_{k=0}^na_kb_{n-k}\right)_{n\in\mathbb{N}}\tag{2} $$ gives us that $$ \left(\sum_{n=0}^\infty\binom{1/2}{n}X^n\right)^2=1+X\tag{3} $$