The problem asks to determine whether $$ \sum_{n=2}^\infty \frac{1}{(\log \ n)^n} $$ converges or not. The answer states that it converges because $ \frac{1}{(\log \ n)^n} \lt \frac{1}{2^n} \text{ for } n \gt 9$.
Could anybody explain why Professor Spivak affirms that?
We need to check that $$\sum_{n=2}^\infty \frac{1}{(\log n)^n}$$ is summable or not. Let be $a_n=\frac{1}{(\log n)^n}$. Being $2<e<3$, then $\log n >2$ when ever $n>9$. This gives:
$$\frac{1}{(\log n)^n}<\frac{1}{2^n}\, \text{ for all } n>9. $$ Since $$\sum_{n=2}^\infty\frac{1}{2^n}$$ is convergent and $\frac{1}{(\log n)^n}>0$ for all $n\geq 2$, so by comparison test of series $$\sum_{n=2}^\infty \frac{1}{(\log n)^n}$$ is convergent.