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Let
$L_{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}\left|D_{n}(t)\right| d t \quad(n=1,2,3, \ldots) .$
Prove that there exists a constant $C>0$ such that
$L_{n}>C \log n \quad(n=1,2,3, \ldots),$
or, more precisely, that the sequence
$\left\{L_{n}-\frac{4}{\pi^{2}} \log n\right\}$
is bounded.
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Solution. We observe that
\begin{aligned} L_{n}=& \frac{1}{\pi} \int_{0}^{\frac{2 \pi}{2 n+1}} \frac{\sin \left(n+\frac{1}{2}\right) t}{\sin \frac{1}{2} t} d t+\\ &+\sum_{k=1}^{n-1} \frac{1}{\pi} \int_{\frac{2 \pi k}{2 n+1}}^{\frac{2 \pi(k+1)}{2 n+1}} \frac{(-1)^{k} \sin \left(n+\frac{1}{2}\right) t}{\sin \frac{1}{2} t} d t+\frac{1}{\pi} \int_{\frac{2 n \pi}{2 n+1}}^{\pi} \frac{(-1)^{n} \sin \left(n+\frac{1}{2}\right) t}{\sin \frac{1}{2} t} d t \end{aligned}
The substitution $u=\left(n+\frac{1}{2}\right) t$ changes the first and last terms into the sum
$\frac{1}{\pi} \int_{0}^{\pi} \frac{\sin u}{\left(n+\frac{1}{2}\right) \sin \left(\frac{u}{2 n+1}\right)} d u+\int_{n \pi}^{\left(n+\frac{1}{2}\right) \pi} \frac{(-1)^{n} \sin u}{\left(n+\frac{1}{2}\right) \sin \left(\frac{u}{2 n+1}\right)} d u .$
The first of these terms tends to $\frac{1}{2 \pi} \int_{0}^{\pi} \sin u d u=\frac{1}{\pi}$ as $n \rightarrow \infty$. The second tends to 0 (for $u \in\left[n \pi,\left(n+\frac{1}{2}\right) \pi\right]$ we have $\sin \left(\frac{u}{2 n+1}\right) \geq \sin \frac{n \pi}{2 n+1}$, which tends to 1 as $n \rightarrow \infty)$

Solution. We observe that
\begin{aligned} L_{n}=& \frac{1}{\pi} \int_{0}^{\frac{2 \pi}{2 n+1}} \frac{\sin \left(n+\frac{1}{2}\right) t}{\sin \frac{1}{2} t} d t+\\ &+\sum_{k=1}^{n-1} \frac{1}{\pi} \int_{\frac{2 \pi k}{2 n+1}}^{\frac{2 \pi(k+1)}{2 n+1}} \frac{(-1)^{k} \sin \left(n+\frac{1}{2}\right) t}{\sin \frac{1}{2} t} d t+\frac{1}{\pi} \int_{\frac{2 n \pi}{2 n+1}}^{\pi} \frac{(-1)^{n} \sin \left(n+\frac{1}{2}\right) t}{\sin \frac{1}{2} t} d t \end{aligned}
The substitution $u=\left(n+\frac{1}{2}\right) t$ changes the first and last terms into the sum
$\frac{1}{\pi} \int_{0}^{\pi} \frac{\sin u}{\left(n+\frac{1}{2}\right) \sin \left(\frac{u}{2 n+1}\right)} d u+\int_{n \pi}^{\left(n+\frac{1}{2}\right) \pi} \frac{(-1)^{n} \sin u}{\left(n+\frac{1}{2}\right) \sin \left(\frac{u}{2 n+1}\right)} d u .$
The first of these terms tends to $\frac{1}{2 \pi} \int_{0}^{\pi} \sin u d u=\frac{1}{\pi}$ as $n \rightarrow \infty$. The second tends to 0 (for $u \in\left[n \pi,\left(n+\frac{1}{2}\right) \pi\right]$ we have $\sin \left(\frac{u}{2 n+1}\right) \geq \sin \frac{n \pi}{2 n+1}$, which tends to 1 as $n \rightarrow \infty)$

The extremes in these inequalities are both bounded. Hence we will be done if we can show that
$\frac{2}{\pi} \log n-\sum_{k=1}^{n-1} \frac{1}{\left(n+\frac{1}{2}\right) \sin \left(\frac{\pi(k+1)}{2 n+1}\right)}$

remains bounded. To do this, we use the fact that there is a constant $K$ such that
$\left|\frac{1}{\sin x}-\frac{1}{x}\right| \leq K x$
for $0<x \leq \frac{\pi}{2}$. This fact in turn is a consequence of the fact that, by L'Hospital's rule,
$\lim _{x \rightarrow 0} \frac{x-\sin x}{x^{2} \sin x}=-\frac{1}{2}$
We thus have
$\sum_{k=1}^{n-1} \frac{1}{\left(n+\frac{1}{2}\right) \sin \left(\frac{\pi(k+1)}{2 n+1}\right)}=E_{n}+\sum_{k=1}^{n-1} \frac{1}{\left(n+\frac{1}{2}\right)\left(\frac{\pi(k+1)}{2 n+1}\right)}$
where
\begin{aligned} \left|E_{n}\right| \leq K \frac{1}{n+\frac{1}{2}} \sum_{k=1}^{n-1} \frac{\pi(k+1)}{2 n+1}=\\ &=\frac{2 K}{\pi(2 n+1)^{2}} \sum_{k=1}^{n} k+1=\frac{2 K}{\pi(2 n+1)^{2}}\left[\frac{(n+1)(n+2)}{2}-1\right] . \end{aligned}
Since the right-hand side tends to $\frac{K}{4 \pi}$ as $n \rightarrow \infty$, we see that $E_{n}$ remains bounded as $n \rightarrow \infty$. We will be finished if we can show that
$\log n-\sum_{k=1}^{n-1} \frac{1}{k+1}$
remains bounded.
by Diamond (75,025 points)

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