If $a_n\in{\mathbb{R}}_{>0},\forall n\in\mathbb{N}_{>0}$, and $a_1=\frac{1}{2},a_n^n+a_{n+1}^n=1.$ Prove that $$\prod_{k=1}^{n}a_k\sim o\left(\frac{1}{\sqrt{n}}\right).$$
In fact, it is easy to see $\frac{1}{2}\leq a_n<1.$ But the Recurrence is a little complex to caculate the generation function.
Obviously, $$\sqrt{a_n^n\,a_{n+1}^n}\le\frac{a_n^n+a_{n+1}^n}{2}=\frac12,$$ so $$\sqrt{a_n\,a_{n+1}}\le\sqrt[n]{\frac12}=2^{-1/n}.$$ Now $$\prod^n_{k=1}a_k=\sqrt{a_1\,a_n}\,\prod^{n-1}_{k=1}\sqrt{a_k\,a_{k+1}},$$ and this implies $$\prod^n_{k=1}a_k<\sqrt{\frac12}\,2^{-H_{n-1}},$$ where $\displaystyle H_n=\sum^n_{k=1}\frac1k$ are the harmonic numbers. Since $H_n>\ln(n+1)$ (cf. answers to Showing inequality for harmonic series.), we obtain $$\prod^n_{k=1}a_k<\sqrt{\frac12}\,2^{-\ln n}=\sqrt{\frac12}\,n^{-\ln 2}.$$ But $\ln 2>\frac12$, so the RHS is $\displaystyle o\left(\frac1{\sqrt{n}}\right)$.
