Let $\{x\}$ denote the fractional part of $x$. I need an example of a sequence with all positive terms $(a_n^2)_{n\geq 1}$ such that $$\lim_{n\to\infty} \{(-1)^na_n^2\}=1\ \ \ \text{and}\ \ \ \lim_{n\to\infty} a_n^2=\infty$$
If the sequence is convergent to $1$, then the subsequence of even and odd powers will also converge to $1$. So we have$$\lim_{n\to\infty} \{(-1)^{2n}a_{2n}^2\}=1\ \ \ \text{and}\ \ \ \lim_{n\to\infty} \{(-1)^{2n-1}a_{2n-1}^2\}=1$$So we have$$\lim_{n\to\infty} \{a_{2n}^2\}=1\ \ \ \text{and}\ \ \ \lim_{n\to\infty} \{-a_{2n-1}^2\}=1$$Since $a_{n}^2>0,\forall n\in\mathbb{N}$, so we have using $\{-x\}=1-\{x\}$$$\lim_{n\to\infty} \{a_{2n}^2\}=1\ \ \ \text{and}\ \ \ \lim_{n\to\infty} (1-\{a_{2n-1}^2\})=1$$Therefore we obtain$$\lim_{n\to\infty} \{a_{2n}^2\}=1\ \ \ \text{and}\ \ \ \lim_{n\to\infty} \{a_{2n-1}^2\}=0$$
I think that due to the presence of $(-1)^n$, the limit will oscillate and does not exist.Any help will be highly appreciated. Thank you!