On a lower bound for $\sum_{k=1}^{n}{\sigma(k)}$

Question

Let $\sigma(X)$ be the sum of the divisors of $X$.

I want to find bounds for $$\sum_{k=1}^{n}{\sigma(k)}.$$

My Attempt

Since consecutive integers are relatively prime, and since we have the inequality $$\sigma(ab) \leq \sigma(a)\sigma(b)$$ where equality holds if and only if $a$ and $b$ are relatively prime, then we have the following lower bound (by using the Arithmetic Mean-Geometric Mean Inequality) $$\sum_{k=1}^{n}{\sigma(k)} > n\left(\prod_{k=1}^{n}{\sigma(k)}\right)^{1/n} \geq n\left(\sigma\left(\prod_{k=1}^{n}{k}\right)\right)^{1/n} = n\left(\sigma(n!)\right)^{1/n}.$$

My questions are:

(1) Is my derivation of the lower bound for $$\sum_{k=1}^{n}{\sigma(k)}$$ correct?

<p><strong>(2)</strong> Is it possible to do better in <strong>(1)</strong>?</p>

<p><strong>(3)</strong> What is an <em>optimized</em> upper bound for $$\sum_{k=1}^{n}{\sigma(k)}?$$</p>

A very rough estimate is $$\sum_{k=1}^n \sigma (k) \le n^2$$ is this satisfactory enough for you? — Crostul, Feb 25 '16 at 16:18
$\sum_{k=1}^{n}{\sigma(k)} > n\prod_{k=1}^{n}{\sigma(k)}$ - should there be a power/root of $n$ here if you're using AM-GM? — πr8, Feb 25 '16 at 16:18

Jack D'Aurizio · Accepted Answer · 2016-02-26T14:10:44.230

$$\sum_{k=1}^{n}\sigma(k) = \sum_{k=1}^{n}\sum_{d\mid k}d = \sum_{d=1}^{n}d\cdot\left\lfloor\frac{n}{d}\right\rfloor\leq n^2,$$

$$\sum_{k=1}^{n}\sigma(k) \geq \sum_{d=1}^{n} d\cdot\left(\frac{n}{d}-1\right)=\frac{n^2-n}{2}.$$

By refining the same approach (for a reference, you may like Apostol's Introduction to Analytic Number Theory) we also have:

$$ \sum_{k=1}^{n}\sigma(k) = \sum_{q\leq n}\left(\frac{n^2}{2q^2}+\frac{n}{2q}-\frac{n\left\{n/q\right\}}{q}\right)=\frac{\pi^2}{12}\,n^2+C\,n\log n$$

with $\left|C\right|\leq\frac{3}{2}$.

On a lower bound for $\sum_{k=1}^{n}{\sigma(k)}$

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