<s>[INST] Closed form of $\sum\limits_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor^2$? Does $\displaystyle\sum_{i=1}^n\left\lfloor\dfrac{n}{i}\right\rfloor^2$ admit a closed form expression?
 [/INST] To see how the first term in the asymptotic expansion is obtained, put $$a(n) = \sum_{k=1}^n \bigg\lfloor \frac{n}{k} \bigg\rfloor^2$$ and note that
$$a(n+1)-a(n) = 1 + \sum_{k=1}^n 
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor^2 - \bigg\lfloor \frac{n}{k} \bigg\rfloor^2\right) \\=
1 + \sum_{d|n+1 \atop d<n+1} \left(\left(\frac{n+1}{d}\right)^2 - \left(\frac{n+1}{d}-1\right)^2\right)
= \sum_{d|n+1} \left(2\left(\frac{n+1}{d}\right)-1\right) \\=
2\sigma(n+1)-\tau(n+1).$$
It now follows that $$a(n) = 2\sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k) =
\sum_{k=1}^n \left(2\sigma(k)-\tau(k)\right).$$
We can apply the Wiener-Ikehara theorem to this sum, working with the Dirichlet series
$$L(s) = \sum_{n\ge 1} \frac{2\sigma(n)-\tau(n)}{n^s} =
2\zeta(s-1)\zeta(s)-\zeta(s)^2.$$
We have $$\operatorname{Res}(L(s); s=2) = \frac{\pi^2}{3},$$
so that by the aforementioned theorem,
$$a(n) \sim \frac{\pi^2/3}{2} n^2 = \frac{\pi^2}{6} n^2.$$
In fact we can use Mellin-Perron summation to predict, but not quite prove, the next terms in the asymptotic expansion, getting
$$a(n) = \left(\sigma(n)-\frac{1}{2}\tau(n)\right) +
\frac{1}{2\pi i} \int_{5/2-i\infty}^{5/2+i\infty} L(s) n^s \frac{ds}{s}$$
which yields
$$a(n) \sim \left(\sigma(n)-\frac{1}{2}\tau(n)\right)
+\frac{\pi^2}{6} n^2 - (\log n + 2 \gamma)n - \frac{1}{6}.$$
This approximation is quite good, giving $16085.71386$ for $n=100$ when the correct value is $16116$ and $1639203.715$ for $n=1000$ when the correct value is $1639093.$
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