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

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 $$\sin{x} so $$\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

• $$\frac{1}{(k+1)^2} \le \frac{1}{(k+\frac12)(k+\frac32)} = \frac{1}{k+\frac12} - \frac{1}{k+\frac32}$$ – achille hui Sep 18 '13 at 4:22
• What are the two questions before this one (items (1) and (2)), on the document you linked to? Are they preliminary steps? If so, why not including them in your question? – Did Sep 18 '13 at 11:05

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 $$\frac{1}{1+\tfrac{1}{2}}-\frac{1}{n+\tfrac{3}{2}} = \frac{2}{3}-\frac{1}{n+\tfrac{3}{2}}<\frac{2}{3}<\ln(2).$$
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, $$\sum_{k=2}^{n}\frac{1}{k^2}=\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}\right)+\frac{1} {5^2}+...+\frac{1}{n^2}\le 0.4236...+\frac{1}{4\cdot 5}+\frac{1}{5\cdot6}...+\frac{1}{(n-1)\cdot n}=$$ $$=0.4236...+\frac{1}{4}-\frac{1}{n}\le 0.68< \ln 2.$$