show that
$$\sum_{k=1}^{n}\sin\dfrac{1}{(k+1)^2}\le\ln{2}$$
I think this is nice inequality, and idea maybe use this $$\sin{x}<x$$ so $$\sum_{k=1}^{n}\sin{\dfrac{1}{(n+1)^2}}<\sum_{k=1}^{n}\dfrac{1}{(k+1)^2}<\dfrac{\pi^2}{6}-1\approx 0.644<\ln{2}$$
But this problem is from Middle school students compution,so they don't know
$$\sum_{k=1}^{\infty}\dfrac{1}{k^2}=\dfrac{\pi^2}{6}$$ so I think this problem have other nice methods? Thank you
and this problem is from http://tieba.baidu.com/p/2600561301