Stuck with number theory identity about the divisor function

Question

So I have to show that $$\sum_{m=1}^{n} \sigma (m) = \sum_{k=1}^{n} k \cdot \big[\frac{n}{k}\big]$$ where $\sigma$ is the divisor function for x = 1 and [n] is the floor function of n for example [2.4] = 2. I tried something around Indunction to solve this problem but the n in the floor function gives me a hard time doing so. Other tries I did was to rewrite the right part to $$ \sigma_0(n) \cdot n + \sum_{d \nmid n,0<d<n} k \cdot \big[\frac{n}{k}\big] $$ but that doesn't help much either since I dont see or know a relation between $\sigma_0 \text{ and } \sigma_1$. Any tips would be appreciated. I still would like to solve it on my own based on some advice but at the moment I am stuck

Answer 1

We have $\sigma(m)=\sum_{d|m}d$ so $$ \sum_{m=1}^n\sigma(m)=\sum_{m=1}^n\sum_{d|m}d=\sum_{d=1}^nd\sum_{d|m}1=\sum_{d=1}^nd\left\lfloor\frac{n}{d}\right\rfloor $$ because there are $\left\lfloor\frac{n}{d}\right\rfloor$ multiples of $d$ in $[\![1,n]\!]$.

Answer 2

Illustration:

For $n=6$

There's a value of $k$ in the table below iff $k\mid m$. If you add down the columns, you get $\sigma(m)$ for each column. If you read across the rows, you see how often $k$ is used as a divisor.

Then the sum of row sums is equal to the sum of column sums to illustrate your identity.

$$\begin{array}{l|cccccc|c}k\;\:\backslash \;\; m&1&2&3&4&5&6& k\cdot \left\lfloor\frac{n}{k} \right\rfloor \\\hline 1&1&1&1&1&1&1&6\\ 2&&2&&2&&2&6\\ 3&&&3&&&3&6\\ 4&&&&4&&&4\\ 5&&&&&5&&5\\ 6&&&&&&6&6\\\hline \sigma(m)&1&3&4&7&6&12 & \sum\sigma(m)=\sum k\cdot \left\lfloor\frac{n}{k} \right\rfloor \end{array}$$