Prove or disproe:

For any positive integer n and $\displaystyle x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0$

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- Mar 4th 2010, 06:24 PMelimAn Inequality? (A Challenging One: Prove or disprove)
Prove or disproe:

For any positive integer n and $\displaystyle x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0$ - Mar 4th 2010, 10:49 PMSudharaka
- Mar 5th 2010, 07:39 AMelimSome how I just could not see this clearly before
Say n (>1) is the smallest integer such that for some minimal point$\displaystyle x_0 \in (0,\pi)$,

$\displaystyle \sum_{k=1}^n \frac{1}{k} \sin kx_0 \leq 0$,

then $\displaystyle \sin nx_0 < 0$

On the other hand the critical points for the function $\displaystyle f_n(x) = \sum_{k=1}^n \frac{1}{k}\sin kx$ are

$\displaystyle x = \frac{(2m+1)\pi}{n+1}, \quad m=0,\cdots,\left[\frac{n-1}{2}\right]$

They are the points in $\displaystyle (0,\pi)$ such that

$\displaystyle \sum_{k=1}^n \cos kx = \frac{\cos \frac{(n+1)x}{2} \sin \frac{nx}{2}}{\sin \frac{x}{2}} = 0$

(note that $\displaystyle \sin \frac{nx_0}{2} \neq 0$ by our finding above)

But $\displaystyle \sin \frac{n(2m+1)\pi}{n+1} = \sin \left( 2m\pi+\frac{n-2m}{n+1}\right) = \sin \frac{n-2m}{n+1} > 0$

Thus we get a contradiction and we actually proved that

$\displaystyle \forall n \in \mathbf{N}^+, \forall x\in (0,\pi) \left( \sum_{k=1}^n \frac{1}{k}\sin kx > 0\right)$

This is a bit stronger statement.

Thanks Sudharaka for the direction. I thought this problem needs more advanced tech to solve. - Mar 5th 2010, 08:19 AMSudharaka
Dear elim,

We have to show that for any positive integer n and $\displaystyle

x \in [0,\pi], \quad \sum_{k=1}^n \frac{1}{k} \sin kx \geq 0$

When n=1,

$\displaystyle \sum_{k=1}^1 \frac{1}{k} \sin kx =sinx\geq{0}~Since,~x \in [0,\pi]$

Therefore the expression is true for n=1

Suppose the expression is true for n=p where $\displaystyle p\in{Z^{+}}$

Then, $\displaystyle \sum_{k=1}^p \frac{1}{k} \sin kx\geq0$

Now, $\displaystyle \frac{1}{(p+1)} \sin (p+1)x\geq{0}~Since,~p+1>0~and~x\in[0,\pi]$

Therefore, $\displaystyle \sum_{k=1}^p \frac{1}{k} \sin kx+\frac{1}{(p+1)} \sin (p+1)x\geq{0}$

Hence, $\displaystyle \sum_{k=1}^{p+1} \frac{1}{k} \sin kx\geq0$

Therefore our the expression is true for $\displaystyle n=p+1~where~p\in{Z^+}$

Hence by mathamatical induction the expression is true for all $\displaystyle n\in{Z^+}$

Hope this will help you. - Mar 5th 2010, 10:29 AMelim
This is not true. And is the hard part.

On the other hand, actually I did used math induction to prove the inequality is true for all positive integer n and all $\displaystyle x \in [0,\pi]$ by showing that no such an integer n>1 and $\displaystyle x_0 \in (0,\pi)$ to let

$\displaystyle \sum_{k=1}^n \frac{1}{k} \sin kx_0 \leq 0$ happen assuming

(**) $\displaystyle \sum_{k=1}^m \frac{1}{k} \sin kx_0 > 0$ for $\displaystyle 1 \leq m < n, \quad x \in (0,\pi)$ which is true for m=1