Answer
Diverges
Work Step by Step
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.
