Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{1}{1+2+3+....n}=\dfrac{2}{n(n+1)}$ and $b_n=\dfrac{1}{ n^2}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{2}{n(n+1)}}{\dfrac{1}{ n^2}}$
or, $ =\lim\limits_{n \to \infty} \dfrac{2n}{n+1}$
and $\lim\limits_{n \to \infty} \dfrac{2}{1+1/n}=2 \ne 0 \ne \infty $
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ is a convergent p-series with $p \gt 1$
Thus, the series converges by the limit comparison test.
