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Let $0<x<\pi$. $n$ be a natural number.

How to prove

$$\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots+ \frac{\sin nx}{n}>0$$

Michael Rozenberg
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noot
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    Hint: Try by induction. – Ivan Di Liberti Aug 23 '13 at 08:43
  • could you explain – noot Aug 23 '13 at 08:44
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    http://math.stackexchange.com/questions/175768/show-1-2-sum-n-1n-cos-n-x-frac-sin-n-1-2-x-sin-fracx2 – Ivan Di Liberti Aug 23 '13 at 09:00
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    @Ivan: Induction? How? – Did Aug 23 '13 at 09:33
  • Well you can easily prove this: http://math.stackexchange.com/questions/175768/show-1-2-sum-n-1n-cos-n-x-frac-sin-n-1-2-x-sin-fracx2?lq=1 by induction, so i though you could prove that by induction, maybe I am wrong. – Ivan Di Liberti Aug 23 '13 at 10:00
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    @Ivan - In the link you assume the identity for $n-1$ and use trigonometric identities to add to the $n-1$ terms and show that that the identity is true for $n$ terms (hence for all natural numbers). But in this question the induction hypothesis will only tell you that the sum of the $n-1$ terms is positive - but with no knowledge of how much, after adding a negative term to it (the $n$-th term may be negative) you wouldn't know if the sum is now positive or negative – Belgi Aug 23 '13 at 10:18
  • The Maple command $evalc(Re(sum(sin(k*x)/k, k = 1 .. infinity)))$ produces $$-1/2,\arctan \left( -\sin \left( x \right) ,-\cos \left( x \right) +1 \right) +1/2,\arctan \left( \sin \left( x \right) ,-\cos \left( x \right) +1 \right) . $$ See ?arctan for info. – user64494 Aug 23 '13 at 11:24
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    It would be helpful to take a look at the book of Polya and Szego(i.e., the Problems and Theorems in Analysis II Chapter 6, Section 3, No.23-25) – zy_ Aug 23 '13 at 13:36
  • to y zhao could you explain, I do not have that book – noot Aug 23 '13 at 13:49
  • They claim and prove that i)The sum $$A(n,x):=\sum_{k=1}^{k=n}\frac {\sin kx} k$$ has the only maximums at $x=\frac \pi {n+1}(2j-1),\ j=1\dots,[\frac {n+1} 2] $ and the only minimums at $x=\frac {2\pi k} {n},,k = 1 \dots , [\frac {n+1} 2]$ on the interval $[0,\pi]$; ii)The maximum value of $A(n,x)$ on $[0.\pi]$ is reached at $x=\frac \pi {n+1} $; iii) $$ \lim_{n \to \infty}A\left(n,\frac \pi {n+1}\right)=\int_0^\pi \frac {\sin(x)} x , dx.$$ – user64494 Aug 23 '13 at 15:07
  • thank you.Then the only extreme values that matter are the minimums. – noot Aug 23 '13 at 15:23
  • how can we derive that the minimums are all positive? – noot Aug 23 '13 at 15:24
  • Look in Polya and Szego(i.e., the Problems and Theorems in Analysis II Chapter 6, Section 3, No.23-25). – user64494 Aug 23 '13 at 15:29
  • I do not have that book. – noot Aug 23 '13 at 15:35
  • Could you explain or send me a scan – noot Aug 23 '13 at 15:35
  • Did the book mention how to prove that the minimums are all positive? – noot Aug 23 '13 at 15:43
  • It has been solved in Math.SE, see the following link:

    http://math.stackexchange.com/questions/376273/inequality-sum-1-le-k-le-n-frac-sin-kxk-ge-0?lq=1

     – zy_ Aug 24 '13 at 04:32

1 Answers1

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Let $f_n(x)=\sum\limits_{k=1}^n\frac{\sin{kx}}{k}$.

For $n=1$ it's obvious.

Let $f_n(x)>0$ for any $0<x<\pi.$

It's enough to prove that $f_{n+1}(x)>0.$

Indeed, let $f_{n+1}(x)\leq0$ for some value of $x\in(0,\pi)$.

Since $f_{n+1}$ is a continuous function on $[0,\pi],$ $f_{n+1}(0)=f_{n+1}(\pi)=0,$

there is $x_0\in(0,\pi)$ for which $f_{n+1}$ gets a minimal value and we know that $f_{n+1}(x_0)\leq0.$

Now, $$f'_{n+1}(x_0)=0$$ or $$\sum_{k=1}^{n+1}\cos{kx_0}=0$$ or $$\sum_{k=1}^{n+1}2\sin\frac{x_0}{2}\cos{kx_0}=0$$ or $$\sum_{k=1}^{n+1}\left(\sin\left(kx_0+\frac{x_0}{2}\right)-\sin\left(kx_0-\frac{x_0}{2}\right)\right)=0$$ or $$\sin\left(nx_0+\frac{3x_0}{2}\right)=\sin\frac{x_0}{2},$$ which gives: $$0\geq f_{n+1}(x_0)=f_n(x_0)+\frac{1}{n+1}\sin(n+1)x_0=$$ $$=f_n(x_0)+\frac{1}{n+1}\sin\left(\left(n+\frac{3}{2}\right)x_0-\frac{x_0}{2}\right)=$$ $$=f_n(x_0)+\frac{1}{n+1}\left(\sin\left(n+\frac{3}{2}\right)x_0\cos\frac{x_0}{2}-\cos\left(n+\frac{3}{2}\right)x_0\sin\frac{x_0}{2}\right)=$$ $$=f_n(x_0)+\frac{\sin\frac{x_0}{2}}{n+1}\left(\cos\frac{x_0}{2}-\cos\left(n+\frac{3}{2}\right)x_0\right)=$$ $$=f_n(x_0)+\frac{\sin\frac{x_0}{2}}{n+1}\left(\left|\cos\left(n+\frac{3}{2}\right)x_0\right|-\cos\left(n+\frac{3}{2}\right)x_0\right)\geq f_n(x_0),$$ which is a contradiction and by induction we are done!

Michael Rozenberg
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  • @Martin R Take $x=\frac{\pi}{4(n+1)}$ and see please better my solution. – Michael Rozenberg Sep 06 '19 at 09:33
  • Can down-voter explain us why did you do it? – Michael Rozenberg Sep 06 '19 at 09:41
  • Since I promised to check it: I think your proof is right (you're not surprised:) but maybe you can make it cleaner - the lines 6-8 are confusing and unneeded, and you may explain the last equality (where you get the absolute value). Perhaps saying at the beginning that it is by induction might be helpful. – user8268 Sep 06 '19 at 18:14
  • @user8268 Thank you! We need lines $6-8$ because we want to prove that $f_{n+1}$ gets on $(0,\pi)$ a minimal value. Since $(0,\pi)$ is open it's not always true. – Michael Rozenberg Sep 06 '19 at 18:31
  • but... you suppose by contradiction that $f_{n+1}(x)\leq 0$ for some $x\in(0,\pi)$, which does imply that there is a minimum inside - no? – user8268 Sep 06 '19 at 18:44
  • @user8268 Follows, but we need to prove it. We have a continuous on $(0,\pi)$ function. It's not always true. – Michael Rozenberg Sep 06 '19 at 18:56
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    @user8268 Now I see! If $f_{n+1}=0$, so $f'_{n+1}=0$ for $x\in(0,\pi),$ which we want. Tank you! My reasoning in lines 6-8 is true, but not necessary. – Michael Rozenberg Sep 06 '19 at 19:06