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