## Calculus (3rd Edition)

Given $$\sum_{n=1}^{\infty} \frac{1}{n+3}$$ Since $f(n)= \frac{1}{n+3}$, 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}^{R} \frac{1}{n+3} d n \\ &=\left.\lim _{t \rightarrow \infty} \ln |n+3|\right|_{1} ^{t} \\ &=\lim _{t \rightarrow \infty}\left(\ln |t+3|-\ln |4|\right) \\ &=\infty\end{aligned} Thus the series diverges.