I need to compare the asymptotic behavior of two double sums and I am having some trouble with it. Any help or suggestion on how to proceed would be very welcome.
First sum. $$ \sum_{2\le i < j \le k} \frac{1}{i \ln i} \frac{1}{j \ln j} = \frac12 (\ln \ln k)^2 \cdot (1 + o(1))$$
This one is simple. It can be obtained by noticing that $$ \sum_{2\le i \le k} \frac{1}{i \ln i} = \ln \ln k \cdot (1 + o(1)), $$ comparing the series with the integral of its continuous version. To get the double sum above, just notice that it is the upper diagonal of the double sum on $\{2, \ldots, k\}^2$.
Let us introduce the second sum.
Second sum. $$ \sum_{2\le i < j \le k} \frac{1}{i \ln i} \frac{1}{j \ln j} \cdot \frac{1}{1 - \frac{i \ln^2 i}{j \ln^2 j}} = \quad ?$$
Ideally, I would like to prove that this second sum has the same asymptotic behavior as the first one. However, I suspect that this does not hold. I list some arguments for and against the same asymptotic behavior.
Arguments against the same behavior.
- The perturbation factor is greater than $1$, so this sum is certainly greater than the first.
- Worse than this, for $j = i+1$ (close to the diagonal) this factor gets close to infinity.
Arguments in favor of the same behavior.
- For $i \ll j$ (far away from the diagonal) we have that the perturbation tends to $1$.
The question does not seem to have attracted much attention... Anyway, since I was able to answer it for myself, I am posting a solution here.
I prove that the same asymptotic behavior holds also for the second sum. Notice that the second sum is the same as \begin{equation*} \sum_{2\le i < j \le k} \frac{1}{i \ln i} \frac{\ln j}{j \ln^2 j - i \ln^2 i} \end{equation*} Since $j > i$ we can write for $j = i+s$ that \begin{equation*} \frac{\ln j}{j \ln^2 j - i \ln^2 i} \le \frac{\ln j}{j \ln^2 j - i \ln^2 j} = \frac{1}{s \ln (i+s)}. \end{equation*} Notice that if we sum only on the range $i \le \ln k$, we obtain \begin{align*} \sum_{i= 2}^{\ln k} \frac{1}{i \ln i} \sum_{s=1}^{k-i} \frac{1}{s \ln (i+s)} &\le \sum_{i= 2}^{\ln k} \frac{1}{i \ln i} \sum_{s=1}^{k-i} \frac{1}{s \ln s} \ \le \sum_{i= 2}^{\ln k} \frac{\ln \ln k}{i \ln i} \\ &\le (\ln \ln k) (\ln \ln \ln k), \end{align*} which is $o\bigl((\ln \ln k)^2\bigr)$. A second observation is that on the range $1 \le s \le i$ we have \begin{equation*} \sum_{i= 2}^{k} \frac{1}{i \ln i} \sum_{s=1}^{i} \frac{1}{s \ln (i+s)} \le \sum_{i= 2}^{k} \frac{1}{i \ln^2 i} \sum_{s=1}^{i} \frac{1}{s} \le \sum_{i= 2}^{k} \frac{1}{i \ln i} \le \ln \ln k. \end{equation*} Finally, for $s > i > \ln k$ we can write $\ln (1 + \tfrac{i}{s}) \geq \tfrac{i}{2s}$ and thus \begin{align*} \sum_{s=i+1}^{k-i} \frac{1}{s \ln (i + s)} &\le \sum_{s=i+1}^{k-i} \frac{1}{s \ln s + i/2} \le \sum_{s=i+1}^{k} \frac{1}{s \ln s} (1 + O(\ln^{-1} k)) \\ &\le (1 + o(1)) [\ln \ln k - \ln \ln i], \end{align*} implying that \begin{align*} \sum_{i = i_0}^{k} \frac{1}{i \ln i} \sum_{s=1}^{k-i} \frac{1}{s \ln (i+s)} &\le o\bigl((\ln \ln k)^2\bigr) + \sum_{i=\ln k}^{k} \frac{(1 + o(1))}{i \ln i} [\ln \ln k - \ln \ln i] \\ &\le (1+o(1)) \left[ (\ln \ln k)^2 - \sum_{i=\ln k}^{k} \frac{\ln \ln i}{i \ln i} \right]. \end{align*} We finish the proof by noticing that we can compare the sum above with the integral of $\tfrac{\ln \ln x}{x \ln x}$, whose primitive is $\tfrac{1}{2} (\ln \ln x)^2$.