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".)



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


