Estimate for $\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}$, where $p$ is a large prime

Question

Is this estimate true? Can anyone give a proof of it?

$$ \sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p)\qquad (p\text{ prime, } p\to\infty) $$
where $ (ab)_p\equiv ab\;(\operatorname{mod}p)$, $0<(ab)_p<p$.

Answer 1

Not an answer but a simple heuristic argument: if you set $r=(ab)_p$, the OP's sum is equal to $$\sum_{1\le r\le p-1}\dfrac{1}{r}\sum_{1\le a\le p-1}\dfrac{1}{a}(ra^{-1})_p$$ This proves immediately that the sum is less than $(p-1)H_{p-1}^2$, asymptotically $p\log(p)^2$, and if we assume (heuristic part) that $(ra^{-1})_p$ has average $(p-1)/2$ we indeed obtain a guess of $p\log(p)^2/2$. Maybe this last part can be made rigorous.

EDIT: if you consider the much simpler SINGLE sum $S(p)=\sum_{1\le a\le p-1}\dfrac{(a^{-1})_p}{a}$, the same heuristic would give an asymptotic of $p\log(p)/2$. However, numerically $S(p)/(p\log(p))$ does NOT seem to tend to a limit, but oscillates between something like $0.38$ and $0.52$. This should be much easier to analyze, and perhaps indicate that there is also some oscillation in the OP's original question, with no limit.

Answer 2

Not an answer, but an argument that your sum is between $(1/4+o(1))p\log^2(p)$ and $(3/4+o(1))p\log^2(p)$.

Write $S(p)$ for your sum. Separate the sum into pieces according to the integer part of $(ab/p)$: $$ \begin{align*} S(p)&=\sum_{a=1}^{p-1}\frac{1}{a}\sum_{k=0}^{a-1} \sum_{kp/a<b<(k+1)p/a} \frac{b}{ab-kp}\\ &\sim\sum_{a=1}^{p-1}\frac{1}{a}\sum_{k=0}^{a-1} \frac{kp}{a}\sum_{kp/a<b<(k+1)p/a} \frac{1}{ab-kp}.\\ \end{align*} $$ The innermost sum is the sum of reciprocals of integers in an arithmetic progression. The first term in the progression is $(-pk)_a$, and the sum of the reciprocals of the other terms in the progression is $\log(p/a)/a+O(1/a)$, so $$ \begin{align*} S(p)&\sim\sum_{a=1}^{p-1}\frac{1}{a}\sum_{k=0}^{a-1} \frac{kp}{a}\left[\frac{1}{(-pk)_a}+\frac{1}{a}\log\left(\frac{p}{a}\right)\right]\\ &=p\sum_{a=1}^{p-1}\frac{1}{a^3}\log(p/a)\sum_{k=0}^{p-1}k+p\sum_{a=1}^{p-1}\frac{1}{a^2}\sum_{k=1}^{a-1}\frac{k}{(-pk)_a}\\ &=\frac{1}{4}p\log^2(p)+p\sum_{a=1}^{p-1}\frac{1}{a^2}\sum_{k=1}^{a-1}\frac{k}{(-pk)_a}. \end{align*} $$ This gives the claimed lower bound for $S(p)$. For the upper bound, we observe that for fixed $a$, we have $\{(-pk)_a:1\leq k\leq a-1\}= \{1,\ldots,a-1\}$. So an upper bound for the second term above is $$ p\sum_{a=1}^{p-1}\frac{1}{a^2}\sum_{k=1}^{a-1}\frac{k}{a-k}\sim \frac{1}{2}p\log^2(p). $$

Answer 3

Not really an answer, but here is the plot for the first 200 primes:

Maple code:

with(plots):
f := proc(p) 
 option remember;
 return evalf(add(add(b/a/modp(a*b,p),b=1..p-1),a=1..p-1)/p/ln(p)^2);
end:
listplot([seq(f(ithprime(k)),k=10..200)],style=point);

(Obviously this is completely unintelligent, and much more efficient methods are doubtless possible.)

Answer 4

I feel obliged to flesh out my comments (and to modify my wrong answer, thanks to GH from MO).

Write $n=ab$, and let $H_{k}$ the $k$-th harmonic number, $\tau(n)$ the number of positive divisors of $n$. LHS is greater than the related sum $\sum_{a=1}^{p-1}\sum_{b=1}^{a}\dfrac{b}{a(ab)_{p}}$, that we'll denote by $S$.

Provided the limit as $\tau(n)$ tends to $\infty$ of $f(n):=\frac{2}{\tau(n)}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{d^{2}}{n}$ exists and equals a positive constant $M$, we have:

$S=\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{d^2}{n(n)_{p}}\sim\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{M}{(n)_{p}}\sim\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}M\dfrac{H_{p-1}}{p-1}\sim M\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{\log(p-1)}{p-1}$.

The idea is to consider all divisors of $n$ less than its square root, and to replace $\dfrac{1}{n_p}$ by its average value, which is $\dfrac{H_{p-1}}{p-1}\sim\dfrac{\log(p-1)}{p-1}$ (in my comment I erroneously took the reciprocal of the average value and not the average value of the reciprocals, hence the missing $\log(p-1)$).

We thus obtain $S\sim M\dfrac{\log(p-1)}{p-1}\sum_{n=1}^{(p-1)^2}\dfrac{\tau(n)}{2}$. As $D(x):=\sum_{n=1}^{x}\tau(n)$ is provably asymptotic to $x(\log x+K)$ where $K$ is a positive constant (see Dirichlet divisor problem on Wikipedia), we end up with:

$S\sim\frac{M}{2}\frac{\log(p-1)}{p-1}(p-1)^2(\log (p-1)^2+K)\sim Mp\log^{2} p$ which provides a lower bound for the sum of the OP of the desired order of magnitude.

Edit: the following link: http://www.les-mathematiques.net/phorum/read.php?5,1967230,1967504#msg-1967504 shows that if $M$ exists, it equals $\frac{1}{2}$.