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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.
in Mathematics by Diamond (75,025 points) | 103 views

1 Answer

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Best answer
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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