Is there an answer?

The sum $\sum\limits_{n=2}^{10} \frac1{(\ln(n))^n}$ equals $3.2425\ldots$ by direct computation. The remainder satisfies the following inequality: $\sum_{n=11}^\infty \frac1{(\ln(n))^n} < \sum_{n=11}^\infty \frac1{2^n} < \frac1{2^{10}} < 0.001,$ so this is good enough to determine the value of the infinite sum to three decimal places: $\fbox{3.242}.$