Matematické Fórum

\documentclass{article}
\usepackage{amsmath,amsfonts}
\usepackage{MnSymbol}
\def\N{\mathbb N}
\def\dx{\textnormal dx}
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\begin{document}
It is easy to see that
\[
\left\{(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\cdot\frac{1}{x}\right\}
 =\begin{cases}
  n-\tfrac{1}{x}    & x\in\left (\frac{1}{n},\frac{1}{n-1}\right ),\quad n\in\N ,\text{$n$ even},\\[2mm]
  \tfrac{1}{x}-(n-1)& x\in\left (\frac{1}{n},\frac{1}{n-1}\right ),\quad n\in\N ,\text{$n$ odd, $n\ge 3$}.
  \end{cases}
\]

Therefore,
\begin{align*}
I&=\int_{0}^{1}\left\{(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\cdot\frac{1}{x}\right\}\dx\\[2mm]
%-----
 &=\lim_{N\to\infty}\sum_{n=1}^{N}\int_{\frac{1}{n+1}}^{\frac{1}{n}}
   \left\{(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\cdot\frac{1}{x}\right\}\dx\\[2mm]
%-----
 &=\lim_{N\to\infty}\left (
   \sum_{\substack{n=2\\\text{$n$ even}}}^{2\left\lfloor\frac{N}{2}\right\rfloor}
   \int_{\frac{1}{n}}^{\frac{1}{n-1}}\left (n-\frac{1}{x}\right )\dx
   %
  +\sum_{\substack{n=3\\\text{$n$ odd}}}^{2\left\lfloor\frac{N-1}{2}\right\rfloor +1}
   \int_{\frac{1}{n}}^{\frac{1}{n-1}}\left (\frac{1}{x}-(n-1)\right )\dx
   \right )\\[2mm]
%-----
 &=\lim_{N\to\infty}\left (
   \sum_{\substack{n=2\\\text{$n$ even}}}^{2\left\lfloor\frac{N}{2}\right\rfloor}
   \bigl [nx-\ln (x)\bigr ]_{\frac{1}{n}}^{\frac{1}{n-1}}
   %
  +\sum_{\substack{n=3\\\text{$n$ odd}}}^{2\left\lfloor\frac{N-1}{2}\right\rfloor +1}
   \bigl [\ln (x)-(n-1)x\bigr ]_{\frac{1}{n}}^{\frac{1}{n-1}}
   \right )\\[2mm]
%-----
 &=\lim_{N\to\infty}\left (
   \sum_{\substack{n=2\\\text{$n$ even}}}^{2\left\lfloor\frac{N}{2}\right\rfloor}
   \left (\frac{1}{n-1}-\ln\left (\frac{n}{n-1}\right )\right )
   %
  +\sum_{\substack{n=3\\\text{$n$ odd}}}^{2\left\lfloor\frac{N-1}{2}\right\rfloor +1}
   \left (\ln\left (\frac{n}{n-1}\right )-\frac{1}{n}\right )
   \right )\\[2mm]
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 &=\lim_{N\to\infty}\left (
   \sum_{\substack{n=2\\\text{$n$ even}}}^{2\left\lfloor\frac{N}{2}\right\rfloor}\frac{1}{n-1}
  -\sum_{\substack{n=3\\\text{$n$ odd}}}^{2\left\lfloor\frac{N-1}{2}\right\rfloor +1}\frac{1}{n}
   \right )
   %
  +\lim_{N\to\infty}\left (
   \sum_{\substack{n=2\\\text{$n$ even}}}^{2\left\lfloor\frac{N}{2}\right\rfloor}\ln\left (\frac{n-1}{n}\right )
  +\sum_{\substack{n=3\\\text{$n$ odd}}}^{2\left\lfloor\frac{N-1}{2}\right\rfloor +1}\ln\left (\frac{n}{n-1}\right )
   \right )\\[2mm]
%-----
 &=1
  +\lim_{N\to\infty}
   \ln\left (
   \prod_{n=1}^{\left\lfloor\frac{N}{2}\right\rfloor}\frac{2n-1}{2n}\cdot
   \prod_{n=1}^{\left\lfloor\frac{N-1}{2}\right\rfloor}\frac{2n+1}{2n}
   \right )\\[2mm]
%-----
 &=1
  +\lim_{N\to\infty}
   \ln\left (
   \prod_{n=1}^{\left\lfloor\frac{N}{2}\right\rfloor}\frac{(2n-1)(2n+1)}{(2n)^2}\right )\\[2mm]
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 &=1+\ln\left (\frac{2}{\pi}\right ),
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\end{align*}
where we used the well-known Wallis' product.
\end{document}