For this question, I'm stuck on finding a constant that will make the series convergent. Here is what I have so far. Can anyone please help me out?
Find all positive values of the constant b for which the series $\sum_{n=1}^{\infty} b^{ln(n)}$ is convergent.
|b| has to be less than one in order for the series to be convergent.
$\sum_{n=1}^{\infty} b^{ln(n)} = \frac{1}{1-b}$
$ 1 = \frac{1}{1-b}$
$1(1-b)= 1$
$b=2$
$\sum_{n=1}^{\infty} b^{ln(n)} =\sum_{n=1}^{\infty} e^{\ln (b)ln(n)} =\sum_{n=1}^{\infty} n^{\ln (b) } =\sum_{n=1}^{\infty} \dfrac1{n^{-\ln (b) }} $ which converges for $-\ln(b) > 1$ or $\ln(b) < -1$ or $b < 1/e$.