<s>[INST] 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!
 [/INST] As you properly wrote$$x_n=\sum_{k=1}^n \frac{1}{1^2+2^2+\dots+k^2}=\sum_{k=1}^n \frac{6}{k (k+1) (2 k+1)}$$ Using partial fraction decomposition, as @Ben Grossmann already did it,
$$x_n=\sum_{k=1}^n \frac{6}{k}+\sum_{k=1}^n \frac{6}{k+1}-\sum_{k=1}^n \frac{24}{2 k+1}$$
$$x_n=(6 H_n)+(6 H_{n+1}-6)-(12 H_{n+\frac{1}{2}}-24+24 \log (2))$$ Using the asymptotics
$$x_n=(18-24 \log (2))-\frac{3}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
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