Answer
Divergent
Work Step by Step
Let us consider $f(x)=\dfrac{-2}{n+1}$
Re-write as: $f(x)=-2(\dfrac{1}{n+1})$
Here, the function $f(x)$ is positive, continuous and decreasing for $x \geq 0$
Then $\lim\limits_{x \to \infty} \int_0^\infty \dfrac{1}{ x+1}dx= \lim\limits_{k \to \infty} [\log (x+1)]_{0}^k=\ln (k+1)-\ln 1=\infty$
Thus, the sequence $\Sigma_{n=0}^\infty \dfrac{-2}{ n+1}$ is Divergent.
