Reference Request - Sharp Estimates for a Logarithmic Sum

Question

Can anybody suggest a good (e.g. "non-technical") introduction to estimating bounds for logarithmic sums of the form

$$\sum_{i=1}^{r}{{\alpha_i}{\log(q_i)}}$$

where the $$\alpha_i$$ are positive integers (not necessarily distinct) and the $$q_i$$ are odd primes?

The reason why I ask this question is because I am trying to estimate (both lower and upper) bounds for the logarithm of a number-theoretic function, specifically $$\sigma_{1}(N)$$

I was able to show (in 2008) that

$$\sigma_{1}({q_i}^{\alpha_i}) \le \frac{2}{3}\frac{N}{{q_i}^{\alpha_i}}$$

for all $i = 1, 2, ..., r = \omega(N)$, where $N$ is an OPN (i.e. Odd Perfect Number) and ${q_i}^{\alpha_i} || N$ for all $i$. You will be able to get an upper bound for the logarithmic sum given above, by first taking logarithms of both sides of the inequality, then summing over all $i$.

Unfortunately, for the "numbers" $N$ that I am considering, the current literature (on OPNs) do not point to an approach "strong enough" to prove nonexistence of such "numbers" $N$. (This is because the upper bound alluded to in the previous paragraph is still (of course) dependent on $r = \omega(N)$).

(And that is the reason why --) I'd be particularly interested in an analytic-number-theoretic approach. =) (Thanks to Gerry Myerson for encouraging this "clarification".)

Answer 1

Rather than multiplying, we sum $\forall i \in {1, 2, \ldots \omega(N)}$ to get:

$$\sum_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} \le \frac{2\omega(N)}{3}$$

Following Nielsen, we know that (for lack of an "effective" upper bound for $\omega(N)$):

$$\inf\left({\frac{2\omega(N)}{3}}\right) \ge 6$$

Now multiplying, $\forall i \in {1, 2, \ldots \omega(N)}$ we get:

$$\prod_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} = \frac{{\displaystyle\prod_{j = 1}^{\omega(N)}{{q_j}^{\beta_j}}\displaystyle\prod_{j = 1}^{\omega(N)}}{{\sigma({q_j}^{\beta_j})}}}{\displaystyle\prod_{j = 1}^{\omega(N)}{N}} = \displaystyle\frac{N{\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j}})}}{\displaystyle\prod_{j = 1}^{\omega(N)}{N}} = \displaystyle\frac{\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j}})}{\displaystyle\prod_{j = 1}^{\omega(N) - 1}{N}} = \displaystyle\frac{\sigma(N)}{N^{\omega(N) - 1}}$$

But:

$$\displaystyle\frac{\sigma(N)}{N^{\omega(N) - 1}} = \displaystyle\frac{2N}{N^{\omega(N) - 1}} = \displaystyle\frac{2}{N^{\omega(N) - 2}} = 2{N^{2 - \omega(N)}}$$

Since $7 \le \omega(N) - 2 = r$, it follows that $N^7 \le N^r$, which gives:

$$N^{-r} \le N^{-7}$$

Consequently:

$$\prod_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} \le 2N^{-7} = \displaystyle\frac{2}{N^7}$$

Taking logarithms of both sides of the last inequality:

$$\displaystyle\sum_{j = 1}^{\omega(N)}\log(\displaystyle\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N}) \le \log(2) - 7\log(N)$$

This is as far as I could go, using only elementary notions that are familiar to me. I will let you guys know if I discover anything else in the coming days.

Post-Edit:

I missed it!

$$\displaystyle\sum_{j = 1}^{\omega(N)}\log(\displaystyle\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N}) = \displaystyle\sum_{j = 1}^{\omega(N)}{\left(\log({q_j}^{\beta_j}) + \log(\sigma({q_j}^{\beta_j})) - \log(N)\right)}$$ $$= \displaystyle\sum_{j = 1}^{\omega(N)}{\log({q_j}^{\beta_j})} + \displaystyle\sum_{j = 1}^{\omega(N)}{\log(\sigma({q_j}^{\beta_j}))} - \displaystyle\sum_{j = 1}^{\omega(N)}{\log(N)}$$ $$= \log(\displaystyle\prod_{j = 1}^{\omega(N)}{{q_j}^{\beta_j}}) + \log(\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j})}) - \log(\displaystyle\prod_{j = 1}^{\omega(N)}{N})$$ $$= \log(N) + \log(\sigma(N)) - \log(N^{\omega(N)}) \le \log(2) - 7\log(N)$$

which implies that:

$$8\log(N) + \log(\sigma(N)) - \omega(N)\log(N) \le \log(2)$$

Finally, we have:

$$\log(\sigma_{1}(N)) \le \log(2) + \left(\omega(N) - 8\right)\log(N)$$

Equality holds if and only if $\omega(N) = 9$.