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