How to prove that $\forall x \in \mathbb{R}$,$n \in \mathbb{N}$, we have \begin{align} \sum_{k=1}^{n}\frac{|\sin{kx}|}{k}\ge |\sin{nx}| \end{align} I know that this partial sum will diverge for $x\not = m\pi$, but I don't know how to prove this inequality, I have tried Abel summation, but it doesn't work because I can't give a lower bound for $\sum |\sin{kx}|$. Thanks for your attention.
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|||||||||||||
|
|
You can get a lower bound as follows: $$ |\sin(k\,x)|\ge|\sin(k\,x)|^2=\frac{1-\cos(2\,k\,x)}{2}. $$ From this it is easy to get $$ \sum_{k=1}^n\frac{|\sin(k\,x)|}{k}\ge\sum_{k=1}^n\frac{1}{2\,k}-\sum_{k=1}^n\frac{\cos(2\,k\,x)}{2\,k}\ge\frac12\log n-C(x),\quad x\ne0,\pi, $$ for some $C(x)>0$. This will prove the inequality for large values of $n$. |
|||||||||
|
|
Let's assume $nx\le\pi$. A glance at the graph of $\sin x$ shows that for $0\le k\le n$ the line through the origin and the point $(kx,\sin kx)$ passes through or above the point $(nx,\sin nx)$, so $\sin kx\ge(k/n)\sin nx$. So $$\sum_{k=1}^n{\sin kx\over k}\ge\sum_{k=1}^n{\sin nx\over n}=\sin nx$$ I'm confident that the general case is just a matter of filling in a few details. |
|||||||||||
|
