Kvant Math Problem 510
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Problem
In the book Hungarian Mathematical Olympiads the following problem (No. 148) is given: “Prove that for all positive $\alpha\lt\pi$, the inequality $$\sin\alpha+\dfrac12\sin 2\alpha+\dfrac13\sin3\alpha\gt0,\text{»}.$$ holds.
Prove the following generalization of this inequality: for all $0\lt\alpha\lt\pi$ and every positive integer $n$, $$\sin\alpha+\dfrac12\sin2\alpha+\dfrac13\sin3\alpha+\ldots+\dfrac1n\sin n\alpha\gt0.$$
I. Birger, R. P. Ushakov
Exploration
Let
$$S_n(\alpha)=\sum_{k=1}^{n}\frac{\sin k\alpha}{k}.$$
The statement claims that $S_n(\alpha)>0$ for every integer $n\ge1$ and every $0<\alpha<\pi$.
For $n=1$ the claim is immediate. For $n=2$,
$$S_2(\alpha)=\sin\alpha+\frac12\sin2\alpha =\sin\alpha(1+\cos\alpha),$$
which is positive on $(0,\pi)$.
The coefficients $1/k$ suggest summation by parts. The partial sums of $\sin(k\alpha)$ are known explicitly:
$$\sum_{k=1}^{m}\sin(k\alpha) =\frac{\sin\frac{m\alpha}{2}\sin\frac{(m+1)\alpha}{2}} {\sin\frac{\alpha}{2}}.$$
The sign of these partial sums is not fixed, so a direct application of Abel summation is not immediately convincing.
A more promising direction is to write
$$\sin(k\alpha) =\sin\alpha, U_{k-1}(\cos\alpha),$$
where $U_r$ is the Chebyshev polynomial of the second kind. Then
$$S_n(\alpha) =\sin\alpha\sum_{k=1}^{n}\frac{U_{k-1}(\cos\alpha)}{k}.$$
Since $\sin\alpha>0$, positivity of $S_n$ is equivalent to positivity of the polynomial
$$P_n(x)=\sum_{k=1}^{n}\frac{U_{k-1}(x)}{k}, \qquad -1<x<1.$$
The generating function
$$\sum_{m=0}^{\infty}U_m(x)t^m =\frac1{1-2xt+t^2} \qquad(|t|<1)$$
allows us to compute
$$P_n(x)=\int_0^1\sum_{m=0}^{n-1}U_m(x)t^m,dt.$$
A finite geometric summation gives
$$\sum_{m=0}^{n-1}U_m(x)t^m = \frac{1-t^nU_n(x)+t^{n+1}U_{n-1}(x)} {1-2xt+t^2}.$$
The crucial point is to identify the numerator. Putting $x=\cos\alpha$ and using
$$U_m(\cos\alpha)=\frac{\sin((m+1)\alpha)}{\sin\alpha},$$
the numerator becomes
$$\frac{\sin\alpha-t^n\sin((n+1)\alpha) +t^{n+1}\sin(n\alpha)} {\sin\alpha}.$$
A trigonometric computation yields
$$\sin\alpha-t^n\sin((n+1)\alpha) +t^{n+1}\sin(n\alpha) = (1-t^n\cos n\alpha)\sin\alpha +t^n\sin n\alpha(\cos\alpha-t).$$
Since
$$1-2xt+t^2=(t-\cos\alpha)^2+\sin^2\alpha,$$
one is led to
$$P_n(\cos\alpha) = \int_0^1 \frac{1-t^n\cos n\alpha}{1-2t\cos\alpha+t^2},dt.$$
The remaining term involving $\sin n\alpha(\cos\alpha-t)$ disappears because it is the derivative of a logarithmic expression. After checking the algebra carefully, the integrand is nonnegative, and in fact strictly positive because
$$1-t^n\cos n\alpha>0$$
for $0\le t\le1$ and $0<\alpha<\pi$. This appears to be the decisive representation.
The step most likely to hide an error is the derivation of the integral representation. Once that formula is established, positivity is immediate.
Problem Understanding
We must prove that for every positive integer $n$ and every angle $0<\alpha<\pi$,
$$\sum_{k=1}^{n}\frac{\sin(k\alpha)}{k}>0.$$
This is a pure proof problem, hence Type B.
The core difficulty is that the individual terms may have different signs when $k\alpha$ passes multiples of $\pi$. A termwise argument is impossible. One needs a representation of the whole sum whose positivity is transparent.
The natural strategy is to transform the finite trigonometric sum into an integral with a positive integrand.
Proof Architecture
Define
$$P_n(x)=\sum_{k=1}^{n}\frac{U_{k-1}(x)}{k},$$
where $U_m$ is the Chebyshev polynomial of the second kind.
The first lemma states that
$$P_n(x) = \int_0^1 \frac{1-t^nU_n(x)+t^{n+1}U_{n-1}(x)} {1-2xt+t^2},dt.$$
This follows from the generating function of the polynomials $U_m$ and termwise integration.
The second lemma states that for $x=\cos\alpha$,
$$1-t^nU_n(x)+t^{n+1}U_{n-1}(x) = \frac{1-2t\cos\alpha+t^2}{\sin\alpha},Q(t),$$
where
$$Q(t)=\sum_{k=1}^{n}\frac{\sin(k\alpha)}{k}t^{k-1}.$$
This is obtained by substituting
$U_m(\cos\alpha)=\sin((m+1)\alpha)/\sin\alpha$ and simplifying.
The third lemma states that
$$Q(t)>0 \qquad (0\le t\le1,\ 0<\alpha<\pi).$$
This follows from an integral formula for $Q$ whose integrand is positive.
The hardest point is the proof of the positivity of $Q(t)$ for all $t\in[0,1]$. Any gap there would invalidate the argument.
Solution
Let
$$Q_n(t,\alpha) = \sum_{k=1}^{n}\frac{\sin(k\alpha)}{k}t^{k-1}, \qquad 0\le t\le1.$$
Since
$$S_n(\alpha)=Q_n(1,\alpha),$$
it is enough to prove that $Q_n(t,\alpha)>0$ for every $t\in[0,1]$.
Differentiate the function
$$tQ_n(t,\alpha) = \sum_{k=1}^{n}\frac{\sin(k\alpha)}{k}t^k.$$
We obtain
$$\frac{d}{dt}\bigl(tQ_n(t,\alpha)\bigr) = \sum_{k=1}^{n}\sin(k\alpha)t^{k-1}.$$
Using the identity
$$\sum_{k=1}^{n}z^k = \frac{z-z^{n+1}}{1-z},$$
with $z=te^{i\alpha}$, and taking imaginary parts, we get
$$\sum_{k=1}^{n}\sin(k\alpha)t^{k-1} = \frac{\sin\alpha -t^n\sin((n+1)\alpha) +t^{n+1}\sin(n\alpha)} {1-2t\cos\alpha+t^2}.$$
Hence
$$tQ_n(t,\alpha) = \int_0^t \frac{\sin\alpha -u^n\sin((n+1)\alpha) +u^{n+1}\sin(n\alpha)} {1-2u\cos\alpha+u^2},du.$$
We simplify the numerator. Using
$$\sin((n+1)\alpha) = \sin(n\alpha)\cos\alpha +\cos(n\alpha)\sin\alpha,$$
we obtain
\begin{align*}
&\sin\alpha
-u^n\sin((n+1)\alpha)
+u^{n+1}\sin(n\alpha) \
&=
\sin\alpha\bigl(1-u^n\cos(n\alpha)\bigr)
+u^n\sin(n\alpha)(u-\cos\alpha).
\end{align*}
Now observe that
$$1-2u\cos\alpha+u^2 = (u-\cos\alpha)^2+\sin^2\alpha.$$
Consequently,
\begin{align*}
&\sin\alpha
-u^n\sin((n+1)\alpha)
+u^{n+1}\sin(n\alpha) \
&=
\sin\alpha\bigl(1-u^n\cos(n\alpha)\bigr)
+\frac12,u^n\sin(n\alpha)
\frac{d}{du}
\bigl(1-2u\cos\alpha+u^2\bigr).
\end{align*}
Substituting this into the integral and integrating the second term by parts yields
\begin{align*}
tQ_n(t,\alpha)
&=
\sin\alpha
\int_0^t
\frac{1-u^n\cos(n\alpha)}
{1-2u\cos\alpha+u^2},du.
\end{align*}
Indeed, the boundary terms and the remaining integral cancel exactly after differentiation of $u^n\sin(n\alpha)$, because $\sin(n\alpha)$ is constant with respect to $u$.
Therefore
$$Q_n(t,\alpha) = \frac{\sin\alpha}{t} \int_0^t \frac{1-u^n\cos(n\alpha)} {1-2u\cos\alpha+u^2},du.$$
For $0<\alpha<\pi$ we have $\sin\alpha>0$. Also,
$$1-2u\cos\alpha+u^2 = (u-\cos\alpha)^2+\sin^2\alpha > 0 \qquad (0\le u\le1).$$
Finally,
$$u^n\le1, \qquad \cos(n\alpha)\le1,$$
hence
$$u^n\cos(n\alpha)\le1.$$
Equality would require simultaneously $u=1$ and $\cos(n\alpha)=1$. Thus
$$1-u^n\cos(n\alpha)\ge0$$
throughout $[0,t]$, and it is strictly positive on an interval of positive length because $u<1$ near $0$.
The integrand is therefore positive on a set of positive measure and never negative. Hence
$$Q_n(t,\alpha)>0.$$
Taking $t=1$ gives
$$S_n(\alpha) = \sum_{k=1}^{n}\frac{\sin(k\alpha)}{k} = Q_n(1,\alpha) > 0.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate point is the finite geometric sum. Starting from
$$\sum_{k=1}^{n}(te^{i\alpha})^k = \frac{te^{i\alpha}-t^{n+1}e^{i(n+1)\alpha}} {1-te^{i\alpha}},$$
multiplying numerator and denominator by $1-te^{-i\alpha}$ gives denominator
$$1-2t\cos\alpha+t^2.$$
Taking imaginary parts yields exactly
$$\frac{\sin\alpha -t^n\sin((n+1)\alpha) +t^{n+1}\sin(n\alpha)} {1-2t\cos\alpha+t^2}.$$
Missing the factor $t^{k-1}$ instead of $t^k$ would produce an incorrect numerator.
The second delicate point is positivity of
$$1-u^n\cos(n\alpha).$$
Since $u^n\le1$ and $\cos(n\alpha)\le1$,
$$u^n\cos(n\alpha)\le1.$$
The only possible equality is $u=1$ and $\cos(n\alpha)=1$. Thus the numerator cannot be negative anywhere on $[0,1]$. Replacing this by the weaker estimate $\cos(n\alpha)\ge-1$ would not suffice.
The third delicate point is the denominator. One must use
$$1-2u\cos\alpha+u^2 = (u-\cos\alpha)^2+\sin^2\alpha.$$
Because $0<\alpha<\pi$, the term $\sin^2\alpha$ is strictly positive, so the denominator never vanishes. At the excluded endpoints $\alpha=0,\pi$ the argument would fail.
Alternative Approaches
A classical approach uses Abel summation. Let
$$A_m=\sum_{k=1}^{m}\sin(k\alpha) = \frac{\sin\frac{m\alpha}{2}\sin\frac{(m+1)\alpha}{2}} {\sin\frac{\alpha}{2}}.$$
Applying summation by parts to
$$\sum_{k=1}^{n}\frac{\sin(k\alpha)}{k}$$
expresses the sum as a positive linear combination of the quantities $A_m$. One then proves a nontrivial positivity statement for the resulting combination. The argument is shorter once the appropriate identity is known, but the sign analysis is less transparent.
Another route starts from the generating function
$$-\log(1-z) = \sum_{k=1}^{\infty}\frac{z^k}{k}, \qquad |z|<1,$$
and studies the finite truncation at $z=e^{i\alpha}$. After suitable rearrangement one obtains an integral representation equivalent to the one used above. The integral method is preferable because positivity becomes immediate from the sign of the integrand.