Answer
$$
\sum_{n=1}^{\infty} \frac{n+2}{n+1}
$$
Apply the Integral Test to the given series we obtain
$$
\int_{1}^{\infty}\frac{x+2}{x+1} d x =\infty .
$$
So, the series diverges.
Work Step by Step
$$
\sum_{n=1}^{\infty} \frac{n+2}{n+1}
$$
Apply the Integral Test to the series
$$
\sum_{n=1}^{\infty} \frac{n+2}{n+1}
$$
The function
$$
f(x)=\frac{x+2}{x+1}
$$
is positive and continuous for $x\gt 1$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{-1}{(x+1)^{2}}.
$$
So, $f^{'}(x) \lt 0 $ for $x \geq1$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{1}^{\infty}\frac{x+2}{x+1} d x &=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{x+2}{x+1} d x \\
&=\lim\limits_{t \to \infty } \int_{1}^{t } \frac{(x+1)+1}{x+1} d x \\
&=\lim\limits_{t \to \infty } \int_{1}^{t } \left[ 1+\frac{1}{x+1} \right] d x \\
&=\lim\limits_{t \to \infty } \left[ x+\ln (x+1) \right] _{1}^{t} \\
&=\infty.
\end{aligned}
$$
So, the series diverges.