Inequality involving double summation

Question

I have the inequality

$$e^{-10}\sum_{a=0}^{n-i}a\sum_{b=0}^{n-1}e^{-abn(a-2bn)}<\ln n$$

A table in Mathematica quickly shows that the largest integer $n$ for which the inequality holds is $n=2$. But sadly Mathematica isn't smart enough to Solve the inequality explicitly.

Can someone suggest what steps I might take to solve it algebraically - i.e., to derive an algebraic condition on $n$ for which the inequality holds?

(Note that this similar to a question I asked here - but this inequality is much harder because $-abn(a-2bn)$ is irreducible.)

P.S.: Here is the Mathematica code in case anyone finds it useful:

TableForm[Quiet[Table[
{n, 
 N[Sum[a*Sum[E^((-a)*b*n*
        (a - 2*b*n)), {b, 0, 
       n - 1}], {a, 0, n - 1}]/
   E^11], N[Log[n]]}, {n, 0, 5}]], 
TableHeadings -> {None, {"n", "function", "Log[n]"}}]

Answer 1

I assume that $i$ in the upper bound for the sum is a misprint and there should be $1$ instead. Then the inequality fails for any $n\ge 3$, because in this case a summand with $a=b=n-1$ of the left hand side equals $$ae^{-abn(a-2bn)-10}=(n-1)e^{(n-1)^3n(2n-1)-10}\ge e^n\ge n\ge \ln n.$$