What are the values of r (with r>0) for which the series converges?

Show that ${r}^{\ln} \left(n\right) = {n}^{\ln} \left(r\right)$. Then determine the values of r (with $r > 0$) for which the series ${\sum}_{n = 1}^{\infty} {r}^{\ln} \left(n\right)$ converges.

Mar 10, 2018

$r < \frac{1}{e}$ is the condition for the convergence of ${\sum}_{n = 1}^{\infty} {r}^{\ln} \left(n\right)$

Explanation:

I will just answer the part about the convergence, the first part having been answered in the comments. We can use ${r}^{\ln} \left(n\right) = {n}^{\ln} \left(r\right)$ to rewrite the sum ${\sum}_{n = 1}^{\infty} {r}^{\ln} \left(n\right)$ in the form

${\sum}_{n = 1}^{\infty} {n}^{\ln} \left(r\right) = {\sum}_{n = 1}^{\infty} \frac{1}{n} ^ p , q \quad m b \otimes \left\{f \mathmr{and}\right\} p = - \ln \left(r\right)$

The series on the right is the series form for the famous Riemann Zeta function. It is well known that this series converges when $p > 1$. Using this result directly gives

$- \ln \left(r\right) > 1 \implies \ln \left(r\right) < - 1 \implies r < {e}^{-} 1 = \frac{1}{e}$

The result about the Riemann Zeta functions is very well known, If you want an ab initio answer, you can try the integral test for convergence .