## Calculus (3rd Edition)

Given $$\sum_{n=5}^{\infty} \frac{1}{\sqrt{n-4}}$$ Since $f(n)= \frac{1}{\sqrt{n-4}}$, is positive, decreasing and continuous for $n\geq5$, then by using the integral test \begin{aligned} \int_{5}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{5}^{t} \frac{1}{\sqrt{n-4}}d n \\ &=\left.\lim _{t \rightarrow \infty}\frac{2}{3} (n-4)^{3/2}\right|_{5} ^{t} \\ &=\frac{2}{3}\lim _{t \rightarrow \infty} (t-4)^{3/2}-1 \\ &=\infty\end{aligned} Thus the series diverges.