Let $nx_n \to a \in \mathbb{R}\setminus\{-\infty, 0, \infty\}$. How to find $\lim\limits_{n\to\infty}\sum_{k=1}^{n}{x_k^2}$

Question

Let $nx_n \to a \in \mathbb{R}\setminus\{-\infty, 0, \infty\}$. How to find $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}{x_k^2}$

For example for $x_n=\frac{a}{n}$ we have that $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}{x_k^2}=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}{\frac{a^2}{n^2}}=a^2\frac{\pi^2}{6}$ by Basel problem

Proposition. For any $c>0$ there exists sequence $x_n$ such that $nx_n \to a$ and $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}{x_k^2}=c$. Proof. Let $c=a^2\frac{\pi^2}{6}$ then $x_n=\frac{a}{n}$. Let $c\neq\frac{\pi^2}{6}$ then consider sequences $x_n$ with property that there exists positive integer $N$ such that $x_m=\frac{a}{m}$ for $m>N$. Then $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}{x_k^2}=x_1^2+\dots+x_N^2+\frac{a^2}{(N+1)^2}+\frac{a^2}{(N+2)^2}+\dots=-(\frac{a^2}{1^2}+\dots+\frac{a^2}{N^2})+\frac{a^2}{1^2}+\dots+\frac{a^2}{N^2}+ x_1^2+\dots+x_N^2+\frac{a^2}{(N+1)^2}+\frac{a^2}{(N+2)^2}+\dots=x_1^2+\dots+x_N^2+a^2\frac{\pi^2}{6}-(\frac{a^2}{1^2}+\dots+\frac{a^2}{N^2})$

Therefor

$x_1^2+\dots+x_N^2+a^2\frac{\pi^2}{6}-(\frac{a^2}{1^2}+\dots+\frac{a^2}{N^2})=c$

for some $N$ and $x_1, x_2, \dots, x_N$

Question. Is there a sequence $x_n$ with property that $nx_n \to a \in \mathbb{R}\setminus\{-\infty, 0, \infty\}$ such that $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}{x_k^2}=+\infty$ ?

Answer 1

Changing the value of $x_1$ does not change the hypothesis but it changes the final answer. So it is impossible to find the limit in terms of $a$. Since $x_n{2}<C/n^{2}$ for some constant $C$ the limit cannot be $\infty$.