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I am solving a question and I can't get over this step: proving $$\sin \frac{1}{k} > \frac{1}{k} - \frac{1}{k^2}$$ where $k$ is a positive integer.

I tried using induction, but I failed. One of my friends managed to prove it using derivatives but I am searching for a solution which does not involve calculus or series. Proving this would help me solve a convergence problem.

andu eu
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2 Answers2

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We just have to think about:$$f(x)=\sin x-x+x^2,x\in[0,1].\\$$ $$f'(x)=\cos x-1+2x.\\$$Let$\ g(x)=f'(x),$than$$g'(x)=-\sin x+2.\\$$ When $x\in[0,1]$ $$g'(x)>0.\\$$At this point, $f'(x)(=g(x))$ is monotonically increasing.
So$$f'(x)\geq f'(0)=0,$$if and only if $x=0$,the equal sign holds.
Than we can know $f(x)$ is monotonically increasing.
So$$f(x)\geq f(0)=0,$$if and only if $x=0$,the equal sign holds.
Let x=$\frac{1}{k}>0$,come to the conclusion:$$\sin \frac{1}{k}>\frac{1}{k}-\frac{1}{k^2}.$$

Piquancy
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    That's a nice solution, but I am looking for one which doesn't use derivates. – andu eu Oct 23 '21 at 06:43
  • I have only used derivatives so far for this type of this inequality,and I don't think any other way would be too convenient. – Piquancy Oct 23 '21 at 07:01
  • Proving it for $[\frac{1}{2}, 1]$ is quite easy, but I can't deal with $(0, \frac{1}{2})$ – andu eu Oct 23 '21 at 07:12
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well if you don't want to use calculus ...

from this geometric proof by David Mitra we have $$\sin t\le t\le \tan t$$ Now $$\tan t/2\ge t/2 $$ $$\sin (t/2)\cos(t/2)\ge \frac{t(1-\sin^2(t/2))}{2}$$ $$\to \sin t\ge t(1-\sin^2 (t/2)\ge t(1-t^2/4)=t-t^3/4$$ Using this result we get $$\sin 1/k\ge \frac{1}{k}-\frac{1}{4k^3}$$ But $\frac{1}{k}\le 1$ so obviously $$\frac{1}{k}-\frac{1}{4k^3}\ge \frac{1}{k}-\frac{1}{k^2}$$ we are done!

Albus Dumbledore
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