## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty}\frac{1}{3^{\ln n}}$$ Since $f(n)=\frac{1}{3^{\ln n}} ,\ \ f'(n) =-\frac{\ln \left(3\right)}{n^{\ln \left(3\right)+1}}$, is positive, decreasing and continuous for $n\geq 1$, then by using the integral test \begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{3^{\ln n}}d n \ \ \ \\ &=\lim _{t \rightarrow \infty} \int_{1}^{t} n^{-\ln 3}d n \\ &=n^{-\ln 3-1}\bigg|_{1}^{t} \\ &=t^{-\ln 3-1}-1\\ &=\infty \end{aligned} Thus the series diverges