For which positive real numbers $a$ does $\sum_{n=1}^\infty a^{\log n}$ converge?

Question

I know the answer is $a<1$. I tried the ratio test which was inconclusive, and also the root test which did not work. Any tips on how to proceed?

Any help would be appreciated, thank you

Answer 1

Notice that

$$ a^{\log n} = \exp( \log a \log n) = n^{\log a } $$

Notice that $log a > 0$ iff $a > 1 $. Thus $\sum a^{\log n }$ diverges if $a \geq 1$ and converges when $ log a < -1 $ or when $ a < 1/e $