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I need to prove $\sin(1/n)<1/n$ for all $n \in \Bbb N$ using mathematical induction.

Dont know how to start. Please help!

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18  
Starting is the easy part! Base case: $n=1$ $$\sin \left(1\over 1\right)<1 \, \color{green}\checkmark$$ –  Git Gud Apr 21 '13 at 20:33
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Why did you take it so literally? –  mopdf Apr 21 '13 at 20:44
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Because he is a mathematician. –  Fixed Point Apr 21 '13 at 21:02
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@mopdf Boys will be boys. –  Git Gud Apr 21 '13 at 21:05
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(You can easily show $\sin(x)<x$ for all positive $x$ using the Taylor series for $\sin$.) –  Potato Apr 21 '13 at 21:07
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1 Answer

Because of \[ \sin(x)=x-\frac{x^3}{3!}+ \frac{x^5}{5!}-\frac{x^7}{7!} +\dots \] As $x^n$ is monotone decreasing on $[0,1]$ in respect to $n$. Hence \[ \sin(x)\leq x \] for $x$ in $[0,1]$. As $\sin(x)\leq 1$ the inequality is even true for all $x\in [0,\infty)$.

We can see that when $x-\sin(x)> 0$ holds for $x\in (0,1]$ this implies that $\sin\big(\frac{1}{n}\big) < \frac{1}{n}$ holds for all $n \in \mathbb{N}$. At first we note that for $x=0$ $\sin(x)-x=0$. The derivative of \[ x - \sin(x)\] is $1-\cos(x)$. With the fundamental theoroem of calculus we know that \[ x-\sin(x) = (1-\cos(\xi)) \cdot x \] where $\xi \in (0,x)$. As $\cos (x) \leq 1$ and $x>0$ we have that $x-\sin(x)$ is positive and hence our inequality holds.

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For some reason OP wants or needs to prove this by induction... –  Martin May 17 '13 at 19:46
    
@Martin in the comments he said he like both the induction and the calculus on the problem. But I have no hope for a useful induction proof –  Dominic Michaelis May 17 '13 at 19:50
    
Okay, in that case I would simply develop the Taylor series up to degree 5 and use that the series alternates and the terms are monotonically decreasing. This allows you to avoid the mean value theorem. –  Martin May 17 '13 at 19:54
    
@Martin I made 2 proofs :) –  Dominic Michaelis May 17 '13 at 19:55
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No. And it reads better now. –  Did May 18 '13 at 5:46
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