Reference Request - Sharp Estimates for a Logarithmic Sum 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".)
 A: 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 $\omega(N) = 9$.
