Limited sum

Relevant wiki: Riemann Sums

We can use Riemann sum to solve this problem. We have $\begin{aligned} \lim_{x\to\infty}\frac{\sum_{l\leqslant x}l^n\lfloor\frac{x}{l}\rfloor}{x^{n+1}}&=&\lim_{x\to\infty}\frac{1}{x}\sum_{l\leqslant x}\frac{l^n}{x^n}\left\lfloor\frac{x}{l}\right\rfloor\\ &=&\int^1_0u^n\left\lfloor\frac{1}{u}\right\rfloor\ du\\ &=&\sum^{\infty}_{k=1}\int^{\frac{1}{k}}_{\frac{1}{k+1}}ku^n\ du\\ &=&\sum^{\infty}_{k=1}k\left[\frac{u^{n+1}}{n+1}\right]^{\frac{1}{k}}_{\frac{1}{k+1}}\\ &=&\frac{1}{n+1}\zeta(n+1). \end{aligned}$ Substituting $n=1,$ the limit tends to $\frac{\pi^2}{12}.$

Relevant wiki: Interchanging the summation and integral sign

We have $\begin{aligned} \sum_{\ell \le x} \ell^k \left\lfloor \tfrac{x}{\ell}\right\rfloor & = \; \sum_{\ell \le x} \sum_{m \le \frac{x}{\ell}} \ell^k \; = \; \sum_{m\ell \le x} \ell^k \; = \; \sum_{n \le x} \sum_{\ell|n} \ell^k \\ & = \; \sum_{n \le x} \sigma_k(n) \; = \; \sum_{n \le x} \sum_{d|n} \left(\tfrac{n}{d}\right)^k \; = \; \sum_{d \le x} \sum_{m \le \tfrac{x}{d}}m^k \end{aligned}$ If $k \ge 2$ then $\sum_{m \le \tfrac{x}{d}} m^k \; = \; \tfrac{1}{k+1}\left(\tfrac{x}{d}\right)^{k+1} + O\left(\big(\tfrac{x}{d}\big)^k\right) \hspace{2cm} x \to \infty$ while $\sum_{d \le x} \tfrac{1}{d^{k+1}} \; = \; \zeta(k+1) + O\big(x^{-k}\big) \hspace{2cm} x \to \infty$ which tells us that $\sum_{\ell \le x} \ell^k \left\lfloor \tfrac{x}{\ell}\right\rfloor \; = \; \tfrac{1}{k+1}\zeta(k+1)x^{k+1} + O(x^{k}) \hspace{2cm} x \to \infty$ and hence $\lim_{x \to \infty} x^{-k-1}\sum_{\ell \le x} \ell^k \left\lfloor \tfrac{x}{\ell}\right\rfloor \; = \; \tfrac{1}{k+1}\zeta(k+1)$ The case $k=1$ is slightly more delicate. We still have $\sum_{m \le \tfrac{x}{d}} m \; = \; \tfrac{1}{2}\left(\tfrac{x}{d}\right)^{2} + O\big(\tfrac{x}{d}\big) \hspace{2cm} x \to \infty$ while $\sum_{d \le x} \tfrac{1}{d^{2}} \; = \; \zeta(2) + O\big(x^{-1}\big) \hspace{2cm} x \to \infty$ which tells us that $\sum_{\ell \le x} \ell \left\lfloor \tfrac{x}{\ell}\right\rfloor \; = \; \tfrac{1}{2}\zeta(2)x^{2} + O(x\ln x) \hspace{2cm} x \to \infty$ and hence $\lim_{x \to \infty} x^{-2}\sum_{\ell \le x} \ell \left\lfloor \tfrac{x}{\ell}\right\rfloor \; = \; \tfrac{1}{2}\zeta(2)$ making the answer $\tfrac{1}{12}\pi^2 = \boxed{0.8224670334}$ .

I could able to solve this problem before I learn Riemann sum, but can't solve the bonus problem anyway.

This isn't a fast method but a much primitive method.

First, I test $x=10$ to clarify what is the problem want for.

Let $f(l) = l \lfloor \frac{x}{l} \rfloor$

$\begin {array} {c|c} l & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline f(l) & 10 & 10 & 9 & 8 & 10 & 6 & 7 & 8 & 9 & 10 \end {array}$

Then I saw the pattern,

When $\frac{x}{2} < l \leq x$ , $f(l) = l$

then $\frac{x}{3} < l \leq \frac{x}{2}$ , $f(l) = 2l$ and so on.

Generalize that, I get $\frac{x}{a+1} < l \leq \frac{x}{a}$ , $f(l) = al$

So I draw a graph which satisfied this, because $x$ tend to $\infty$ so the integer result (which made by floor function) will look like linear result.

(This is a example graph for $x=100$ for some $a$ . For x-axis represent $l$ and y-axis represent $f(l)$ )

So I need to calculate the total area under the trapeziums .

Let the area made by $\frac{x}{a+1} < l \leq \frac{x}{a}$ be $P_a$

The area $P_a$ is equivalent as A.M. sequence which the first term is $f(\frac{x}{a+1})=\frac{ax}{a+1}$ , last term is $f(\frac{x}{a})=x$ and the number of terms is $\frac{x}{a}-\frac{x}{a+1}=\frac{x}{a(a+1)}$

You may see which this represented, A.M. Sum, is equal to trapezium area formula $\frac{1}{2}h(a+b)$ , Math is beautiful!

$P_a=\frac{x}{2a(a+1)}(x+\frac{a}{a+1}x)=\frac{2a+1}{2a(a+1)^2}x^2$

After some partition of fraction,

$P_a=x^2 [\frac{1}{2a}- \frac{1}{2(a+1)} +\frac{1}{2(a+1)^2}]$

As what the question ask for,

$lim_{x\to\infty}{\frac{\displaystyle \sum_{l \leq x} l \left\lfloor\frac{x}{l}\right\rfloor}{x^2}}=lim_{x\to\infty}\frac{\displaystyle \sum_{a=1}^{\infty}P_a}{x^2}=\displaystyle \sum_{a=1}^{\infty}\frac{1}{2a}-\frac{1}{2(a+1)}+\frac{1}{2(a+1)^2}$

It's too long, the partition of fraction will be

$\begin{aligned} \sum_{a=1}^{\infty}\left[\frac{1}{2a}-\frac{1}{2(a+1)}\right]+\sum_{a=1}^{\infty}\frac{1}{2(a+1)^2} &=&\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+ \dots \right]+\frac{1}{2}\left[\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots \right]\\ &=& \frac{1}{2}+\frac{1}{2}\left[\frac{\pi^2}{6}-1\right]\\ &=& \boxed{\frac{\pi^2}{12}} \end{aligned}$