Let $\{x\}$ denote the fractional part of $x$. Does there exist a sequence with all positive terms $(a_n)_{n\geq 1}$ such that $$\lim_{n\to\infty} \{(-1)^n a_n\}=1\ \ \ \text{and}\ \ \ \lim_{n\to\infty} \left\{\frac{(-1)^n a_n}{\sqrt{2}}\right\}=1$$
I tried $a_{n}=n+(-1)^{n+1}2^{-n}$. We have $$(-1)^na_{n}= (-1)^nn-1+(1-2^{-n}) $$ Thus $$\{(-1)^na_n\}=1-2^{-n}\to 1$$
Any help will be highly appreciated. Thank you!