Limit of maximum of $f_{n}(x)=\frac{1}{n}(\sin{x}+\sin{(2x)}+\cdots+\sin{(nx)})$

Question

let $$f_{n}(x)=\dfrac{1}{n}(\sin{x}+\sin{(2x)}+\sin{(3x)}+\cdots+\sin{(nx)}),x\in R,n\in N$$

let $$a_{n}=\max_{x\in R}{(f_{n}(x))}$$

Find this limit $$\lim_{n\to\infty}a_{n}$$

My try: since $$\sin{x}+\sin{(2x)}+\sin{(3x)}+\cdots+\sin{(nx)}=\dfrac{\sin{\dfrac{nx}{2}}\sin{\dfrac{(n+1)x}{2}}}{\sin{\dfrac{x}{2}}}$$ so $$f_{n}(x)=\dfrac{\sin{\dfrac{nx}{2}}\sin{\dfrac{(n+1)x}{2}}}{n\sin{\dfrac{x}{2}}}$$ so follow wa have find the $f_{n}(x)$ maximum.But I can't. and I can't find this limit.Thank you very much!

Answer 1

[The proof has been completed following the strategy suggested in the earlier revisions of the answer.]

For $x = \frac{2u}{n}$ the limit of $a_n$ is $\frac{\sin^2 u}u$ and one can take $u$ to maximize this. Let $M$ be the maximum value of $(\sin^2{u})/u$. We have proved that $\limsup a_n \geq M$ and want to show the opposite inequality to prove $\lim a_n = M$.

Lemma 1. If $b_n = \max_y \sin^2 (ny) / (n \sin y)$, then $a_n \leq \max(b_n,\frac{n}{n+1} b_{n+1}$).

Proof: $\sin (ny) \sin((n+1)y) \leq \max \sin^2(ny), \sin^2((n+1)y)$.

Lemma 2. $\limsup a_n \leq \limsup b_n$.

Proof. From lemma $1$, the definition of lim sup, and its invariance under multiplication by factors that converge to $1$.

Lemma 3. In the optimization problem, we can replace $f_n(x)$ by $g_n(x)=\sin^2(nx)/n \sin x$ (whose maxima are $b_n$ instead of $a_n$) if we can prove that $\limsup b_n \leq M$.

Proof. It would show that $\limsup a_n$ is also $\leq M$, by lemma 2.

Lemma 4. The maximum of $g_n(y) = \sin^2(ny)/(n \sin y)$ occurs with $ny \in (0,\pi)$

Proof. There are $n$ values of $y' \in (0,\pi)$ such that $|\sin(ny')|=|\sin ny|$ and $\sin y' > 0$. The smallest of these $n$ values is the unique one in $(0,\pi/n)$ and it also has the smallest value of $|\sin y'|$. For this best choice of $y'$, $g(y') \leq g(y)$ (it is a fraction where we have kept the numerator the same and increased the denominator).

Lemma 5. $\limsup b_n = M$

Proof. The maxima of $g_n(y)$ occur at values of $y$ less than $\pi/n$. But $g_n(y)=(\frac{y}{\sin y})(\frac{\sin^2 (ny)}{ny})$ where the first part converges to $1$ at the maxima and $M$ is defined as the largest possible value of the second part.

Conclusion. $\limsup a_n \leq M$, and therefore $\lim a_n = M$.