Observe that for $n>0$: $$\large \frac{40n}{(n+3)(n+4)(n+5)}< \frac{40n}{n^3}=\frac{40}{n^2},$$ then:
$$S_N=\sum_{n=1}^{N}\frac{40n}{(n+3)(n+4)(n+5)}< \sum_{n=1}^{N}\frac{40}{n^2},$$ so $S_N$ is an increasing sequence bounded above by $K=\sum_{n=1}^{\infty}\frac{40}{n^2}$ (which is convergent with sum $\pi^2/6$). An increasing sequence bounded above converges...