要如何證明1+1/2+1/3+…+1/n-log(n)收斂？

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)=x1−1[x]+1, x0。則 f(x) 恆正。

1nf(x)dx=1nx1dx−1n1[x]+1dx=logn−21+31++n1 

由 f(x) 恆正知上式遞增

而 f(x)=x([x]+1)[x]+1−x1x2，又 11x2dx 

由比較判別法知 1f(x)dx  收斂，故 logn−21+31++1n  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 1+n1ne1+n1n+1 ，n1.
Taking the natural logarithm, we obtain nln1+n11(n+1)ln1+n1 ，
which yields the double inequality 1n+1ln(n+1)−lnnn1.
Applying the one on the right, we find that an−an−1=n1−ln(n+1)+lnn0，for n2,
so the sequence is increasing. Adding the inequalities
1121ln2−ln131ln3−ln21nlnn−ln(n−1)
we obtain 1+21+31++n11+lnn1+ln(n+1).
Therefore, an1, 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 ]