How do you prove that σ(1) + σ(2) + · · · + σ(n) ≤ n^2 for each positive integer n?

Question

If I fix positive integers n and k with 1 ≤ k ≤ n, I need to figure first out for which integers i with 1≤i≤n is k a term in the sum that defines σ(i) and how many such integers i are there (phrased differently: how many times does k occur in the sum σ(1) + σ(2) + · · · + σ(n) if you write each σ(i) as a sum?)? I know the answer depends on n and k but I can't figure out how to do the proof for the original question without first figuring out this step.

Answer 1

Fix a positive integer $n$.

For each positive integer $k$, there are exactly $\lfloor\frac{n}{k}\rfloor$ elements of $\{1,...,n\}$ which are multiples of $k$, hence \begin{align*} \sum_{i=1}^n \sigma(i)&=\sum_{k=1}^n k \left\lfloor\frac{n}{k}\right\rfloor\\[4pt] &\le\sum_{k=1}^n k{\,\cdot\,}\frac{n}{k}\\[4pt] &=\sum_{k=1}^n n\\[4pt] &=n^2\\[4pt] \end{align*}