Answer
Converges
Work Step by Step
Let $u_n=\dfrac{1}{1+2+3+....n}=\dfrac{2}{n(n+1)}$ and $v_n=\dfrac{1}{ n^2}$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{2}{n(n+1)}}{1/n^2}$
$\implies \lim\limits_{n \to \infty} \dfrac{2n}{(n+1)}=\lim\limits_{n \to \infty} \dfrac{2}{1+\dfrac{1}{n}}=2 $
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{(2)}}$ shows a convergent $p$-series with $p \gt 1$.
Hence, the series converges by the limit comparison test.