如題，及如何估計此尤拉常數？

[ 本帖最後由 larson 於 2013-2-26 10:28 PM 編輯 ]

令 \( f(x) = \frac{1}{x} - \frac{1}{[x]+1} \), \( x>0 \)。則 \( f(x) \) 恆正。

\( \begin{aligned}\int_{1}^{n}f(x)dx & =\int_{1}^{n}\frac{1}{x}dx-\int_{1}^{n}\frac{1}{[x]+1}dx\\
& =\log n-\left(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\right)
\end{aligned} \)

由 \( f(x) \) 恆正知上式遞增

而 \( f(x) = \frac{[x]+1-x}{x([x]+1)}\leq\frac{1}{x^{2}} \)，又 \( \int_1^\infty \frac{1}{x^2} dx <\infty \)

由比較判別法知 \( \int_1^\infty f(x)dx \) 收斂，故 \( \log n-\left(\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\right) \) 收斂 (或遞增有上界得收斂)

因此原數列亦收斂

謝謝，還是有些難懂！

Recall the double inequality \( \displaystyle {\Bigg(\; 1+\frac{1}{n} \Bigg)\;}^n<e<{\Bigg(\; 1+\frac{1}{n} \Bigg)\;}^{n+1} \)，\( n\ge 1 \).
Taking the natural logarithm, we obtain \( \displaystyle n \ln \Bigg(\; 1+\frac{1}{n} \Bigg)\;<1<(n+1)\ln \Bigg(\; 1+\frac{1}{n} \Bigg)\; \)，
which yields the double inequality \( \displaystyle \frac{1}{n+1}<\ln(n+1)-\ln n<\frac{1}{n} \).
Applying the one on the right, we find that \( \displaystyle a_n-a_{n-1}=\frac{1}{n}-\ln(n+1)+\ln n>0 \)，for \( n \ge 2 \),
so the sequence is increasing. Adding the inequalities
\( \displaystyle \matrix{1 \le 1, \cr \frac{1}{2}<\ln 2-\ln 1, \cr \frac{1}{3}<\ln 3-\ln 2, \cr ... \cr \frac{1}{n}<\ln n-\ln(n-1),} \)
we obtain \( \displaystyle 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}<1+\ln n<1+\ln(n+1) \).
Therefore, \( a_n<1 \), for all \( n \). We found that the sequence is increasing and bounded, hence convergent.

出自Putnam and Beyond第473頁
http://www.google.com/search?q=P ... chrome&ie=UTF-8

[ 本帖最後由 bugmens 於 2013-3-7 08:54 PM 編輯 ]