lim(n→∞)[(1/1 + 1/2 + 1/3 + …… + 1/n) - ln n] = ?

2007-01-18 1:08 am
lim(n→∞)[(1/1 + 1/2 + 1/3 + …… + 1/n) - ln n] = ?

回答 (2)

2007-01-18 1:42 am
✔ 最佳答案
Actually this is a Euler-Mascheroni constant
The Euler-Mascheroni constant is a mathematical constant, used mainly in number theory. It is defined as the limiting difference between the harmonic series and the natural logarithm:


圖片參考:http://upload.wikimedia.org/math/7/b/7/7b7b33863fa22d4a3f92f0ebde6524c1.png

It is usually denoted by the lowercase Greek letter γ (gamma). Sometimes it is called simply the Euler constant, though it ought not to be confused with e, which is often called Euler's number.
Its approximate value is 0.57721 56649 01532 86060 65120 90082 40243 10421 59335
So
lim(n→∞)[(1/1 + 1/2 + 1/3 + …… + 1/n) - ln n] = γ
You can find the reason in some books about Analytic number theory

2007-01-17 17:47:00 補充:
即是[(1/1 + 1/2 + 1/3 + …… + 1/n) - ln n] 這個數列是收斂的﹐理由請看解析數論方面的書
參考: WIKI
2007-01-20 12:15 am
Actually this is a Euler-Mascheroni constant

The Euler-Mascheroni constant is a mathematical constant, used mainly in number theory. It is defined as the limiting difference between the harmonic series and the natural logarithm:


It is usually denoted by the lowercase Greek letter γ (gamma). Sometimes it is called simply the Euler constant, though it ought not to be confused with e, which is often called Euler's number.

Its approximate value is 0.57721 56649 01532 86060 65120 90082 40243 10421 59335

So

lim(n→∞)[(1/1 + 1/2 + 1/3 + …… + 1/n) - ln n] = γ

You can find the reason in some books about Analytic number theory


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