Limit of sequence involving reciprocal of the sum of the first squares Let $(x_n)$ a sequence defined by $x_n=\sum_{k=1}^n \frac{1}{1^2+2^2+\dots+k^2}.$ What is $\lim_{n\to \infty}x_n$?
My idea: $1^2+\dots+k^2=\frac{k(k+1)(2k+1)}{6}<\frac{k(k+1)(k+2)}{3}$. Thus
$$x_n>3\sum_{k=1}^n\frac{1}{k(k+1)(k+2)}=3\left(\frac 12 -\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{4}\right).$$
I couldn't continue any further. My assumption is that $x_n\to 2$ or something like that. Please help me solve it! Thank you!

Hint: Note that
$$
\frac{1}{k(k+1)(2k+1)} = \frac 1{k+1} + \frac 1k - \frac 4{2k+1}.
$$
So for instance, with $n = 10$, we end up with the sum
$$
\left(1 + 2\cdot \frac12 + \cdots +  2\cdot \frac 1{10} + \frac 1{11} \right) - 4\left(\frac 13 + \frac 15 + \cdots + \frac 1{21} \right) = \\
1 + 2 \left( \frac 12 - \frac 13 + \cdots -\frac 19 + \frac 1{10}\right) + \frac 1{11} - 4\left(\frac 1{11} + \frac 1{13} + \cdots + \frac 1{21} \right).
$$
