Prove or disprove an elementary inequality: $\sin\frac1n>\sin\frac1{2n+1}+\sin\frac1{2n+2},\quad \forall n=1,2,3,\cdots.$

Question

It is easy to prove that $$\frac{1}{n+1}<\frac{1}{2n+1}+\frac{1}{2n+2}<\frac1{n},\quad \forall n=1,2,3,\cdots.$$ Inspired by this inequality, I want to know the following inequalities: $$\sin\frac{1}{n+1}<\sin\frac{1}{2n+1}+\sin\frac{1}{2n+2}<\sin\frac1{n},\quad \forall n=1,2,3,\cdots.$$ What I have done now is： $$\sin\frac{1}{2n+1}+\sin\frac{1}{2n+2}-\sin\frac{1}{n+1} =\sin\frac{1}{2n+1}+\sin\frac{1}{2n+2}-2\sin\frac{1}{2n+2}\cos\frac1{2n+2}$$ $$=\sin\frac{1}{2n+1}-\sin\frac{1}{2n+2}\left(2\cos\frac1{2n+2}-1\right) >\sin\frac{1}{2n+1}-\sin\frac{1}{2n+2}>0.$$ i.e. the LHS is ture. But the same proof technique doesn't apply to the RHS. But the Wolfram mathemtica gives that it seems right for $1\leq n\leq50$,

Any help and hints will welcome!

for real $0 < x < 1$ we know $ x - \frac{x^3}{6} < \sin x < x - \frac{x^3}{6} + \frac{x^5}{120} $ which should work easily for large $n.$ Note $\sin x < x$ also — Will Jagy, Nov 14 '24 at 02:51

score 5 · Accepted Answer · edited Nov 14 '24 at 02:57

5

Use the following lower and upper bound on $\sin(x)$ for $0<x<1$: $$ x \geq \sin(x) \geq x - \frac{x^2}{6}. $$ For instance for the first inequality you should check: $$ \frac{1}{n+1} < \frac{1}{2n+1} - \frac{1}{6(2n+1)^2} + \frac{1}{2n+2} - \frac{1}{6(2n+2)^2}. $$

edited Nov 14 '24 at 02:57

Riemann

11,801

answered Nov 14 '24 at 02:49

Alireza Bakhtiari

112
10

5

should be $x^3 / 6$ not $x^2/6$ – Will Jagy Nov 14 '24 at 02:52
2

Note that for $x<1$ we have that $x^2/6 \geq x^3/6$. This make the algebra a little bit easier without sacrificing the inequality. – Alireza Bakhtiari Nov 14 '24 at 02:55
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Maybe some comment in the answer for why you can use $x^2$ in the formula rather than the expected $x^3.$ A comment is not enough. – Thomas Andrews Nov 14 '24 at 02:57

Ryszard Szwarc · Answer 2 · 2024-11-14T14:38:52.037

3

The first inequality does not require the MacLaurin expansion. It can be proved similarly as in OP $$\sin{1\over 2n+1}+\sin{1\over 2n+2}>2\sin{1\over 2n+2}\\ > 2\sin{1\over 2n+2}\cos{1\over 2n+2}=\sin{1\over n+1}$$

edited Nov 14 '24 at 14:38

answered Nov 14 '24 at 12:03

Ryszard Szwarc

44,675

Thank you for your answer. Actually your second $\geq$ can be $>$. – Riemann Nov 14 '24 at 12:27

score 2 · Answer 3 · answered Nov 14 '24 at 05:02

One thing you could have done, at least for large $n$, is to use Taylor series $$\sin \left(\frac{1}{2 n+1}\right)+\sin \left(\frac{1}{2 n+2}\right)-\sin \left(\frac{1}{ n+1}\right)=\frac{1}{4n^2}+O\left(\frac{1}{n^3}\right)$$ $$\sin \left(\frac{1}{2 n+1}\right)+\sin \left(\frac{1}{2 n+2}\right)-\sin \left(\frac{1}{ n}\right)=-\frac{3}{4n^2}+O\left(\frac{1}{n^3}\right)$$

Prove or disprove an elementary inequality: $\sin\frac1n>\sin\frac1{2n+1}+\sin\frac1{2n+2},\quad \forall n=1,2,3,\cdots.$

3 Answers3