Cauchy criterion for sine series with factorial

Question

$$a_n=\frac{sin1!}{1*2}+\frac{sin2!}{2*3}+\frac{sin3!}{3*4}+...+\frac{sinn!}{n(n+1)}$$ I have this series and I don't understand how to apply the cauchy criterion $$\lvert a_{n+p}-a_n \rvert \lt ε$$ The only result I get is this one: $$\lvert a_{n+p}-a_n \rvert$$=$$\lvert \sum_{k=1}^{n+p} \frac{sink!}{k(k+1)}- \sum_{k=1}^{n} \frac{sink!}{k(k+1)} \rvert$$ What do I do next?What do I apply?

Answer 1

Hint: $|\sin(x)|\le 1$. Estimate $\left|\sum_{k=n+1}^{n+p} \frac{1}{k(k+1)}\right|$.

Answer 2

|$\sum_{k=1}^{n+p}\frac{\sin(k)!}{k(k+1)}-\sum_{k=1}^n\frac{\sin(k!)}{k(k+1)}|=|\sum_{k=n+1}^{n+p}\frac{\sin(k!)}{k(k+1)}|\leq\sum_{k=n+1}^{n+p}\frac{|\sin(k!)|}{k(k+1)}\leq\sum_{k+n+1}^{n+p}\frac{1}{k(k+1)}=\sum_{k=n+1}^{n+p}(\frac{1}{k}-\frac{1}{k+1})=\frac{1}{n+1}-\frac{1}{n+p+1}<\frac{1}{n+1}$

And from here it is easy.