How prove this $\displaystyle\sum_{k=1}^{n}\sin\frac{1}{(k+1)^2}\le\ln{2}$

Question

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

Answer 1

Following @achille hui's comment we can answer the question using a telescoping approach: $$\sum_{k=1}^{n}\sin{\dfrac{1}{(k+1)^2}}\le\sum_{k=1}^{n}\dfrac{1}{(k+1)^2}\le\sum_{k=1}^{n}\dfrac{1}{(k+\tfrac{1}{2})(k+\tfrac{3}{2})}=\sum_{k=1}^{n}\dfrac{1}{(k+\tfrac{1}{2})}-\dfrac{1}{k+\tfrac{3}{2}}.$$

In the right-most summation, the only terms that survive after cancellation are the first and last: $$\sum_{k=1}^{n}\dfrac{1}{(k+\tfrac{1}{2})}-\dfrac{1}{k+\tfrac{3}{2}}=\frac{1}{1+\tfrac{1}{2}}-\frac{1}{n+\tfrac{3}{2}}.$$

The answer follows since $$\frac{1}{1+\tfrac{1}{2}}-\frac{1}{n+\tfrac{3}{2}} = \frac{2}{3}-\frac{1}{n+\tfrac{3}{2}}<\frac{2}{3}<\ln(2).$$

Answer 2

Once you noticed that $\sin x\le x$ you do not need to know the exact value of $\sum_{k=1}^{\infty}\frac{1}{k^2}.$ Instead, you can approximate it by evaluating first few terms and estimating the tail. More precisely, $$\sum_{k=2}^{n}\frac{1}{k^2}=\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}\right)+\frac{1} {5^2}+...+\frac{1}{n^2}\le 0.4236...+\frac{1}{4\cdot 5}+\frac{1}{5\cdot6}...+\frac{1}{(n-1)\cdot n}=$$ $$=0.4236...+\frac{1}{4}-\frac{1}{n}\le 0.68< \ln 2.$$