Answer
$b\lt \frac{1}{e}$ (with $b\gt 0$)
Work Step by Step
$\Sigma_{n=1}^{\infty}b^{ln(n)}=\Sigma_{n=1}^{\infty}n^{lnb}=\Sigma_{n=1}^{\infty}\frac{1}{n^{-lnb}}$
This is a p-series with $p=-lnb$
So if $p=-lnb\gt 1$, then the series will converge.
$$-lnb\gt 1$$$$lnb\lt -1$$$$b\lt e^{-1}$$$$b\lt \frac{1}{e}$$
Hence, $b\lt \frac{1}{e}$ (with $b\gt 0$).
