OK can somebody show me how sequence 1+1/2+1/3+....+1/n-ln(n) converges.Also how did euler compute this constant,how can I compute it?
Have you attempted to find answers to your questions using Google.
There is also an excellent texbook: Amazon.com: Gamma: Exploring Euler's Constant (Princeton Science Library) (9780691099835): Julian Havil, Freeman Dyson: Books
I suggest you try to find and borrow it from a library (perhaps at the institute you study at).
The Euler-MCLaurin summation has been illustrated in...
http://www.mathhelpforum.com/math-he...es-188663.html
... and it can be used for an approximate computation of the so called harmonic numbers...
$\displaystyle H_{n} \sim \gamma + \ln n + \frac{1}{2n}\ \sum_{k \ge 1} \frac{B_{2k}}{2 k\ n^{2 k}}$ (1)
... where $\displaystyle \gamma$ is the Euler's constant and $\displaystyle B_{2k}$ the Bernoulli numbers. A formula like (1) allows great precision if an 'optimal' value of k is choosen. In 1736 Euler used the formula (1) with k=7 ...
$\displaystyle \gamma \sim H_{n} - \ln n - \frac{1}{2 n} + \frac{1}{12\ n^{2}} - \frac{1}{120\ n^{4}} + \frac{1}{252\ n^{6}}- \frac{1}{240\ n^{8}} + \frac{1}{132\ n^{10}}-$
$\displaystyle - \frac{691}{32760\ n^{12}} + \frac{1}{12\ n^{14}}$ (2)
... and n=10 to compute the first sixteen digit of $\displaystyle \gamma$ obtaining...
$\displaystyle H_{10}= 2.9289682539682539...$
$\displaystyle \ln 10 = 2.302585092994045684...$
$\displaystyle \gamma = .5772156649015329...$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
This is illustrated in Apostol's Calculus Vol. I, and I am on this section right now so it would be helpful for me to try and explain it to you. The basics that you need to understand are Leibniz's Rule (also known as The Alternating Series Test).
Let
$\displaystyle a_1=1, \qqaud a_2=\int_1^2 \frac {dx} x, \qquad a_3=\frac 1 2, \qquad a_4=\int_2^3 \frac {dx} x, \hdots$
where, in general,
$\displaystyle a_{2n-1}=\frac 1 n, \qquad a_{2n}=\int_n^{n+1} \frac {dx} x$
We want to apply Leibniz's Rule to the series $\displaystyle \textstyle \sum (-1)^{n-1} a_n $, so we must prove (A) that the series is monotonic decreasing, and (B) that it's limit as n increases without bound is equal to 0.
To verify (A), note that, by the mean value theorem for integration, $\displaystyle \frac 1 c = \int_n^{n+1} \frac {dx} x$ for some $\displaystyle c\in[n,n+1]$. Therefore, $\displaystyle \frac 1 n \ge \frac 1 c\ge \frac 1 {n+1}$, and so $\displaystyle a_{2n-1}\ge a_{2n}\ge a_{2n+1}$. This covers all possibilities, so $\displaystyle \{a_n\}$ is monotonically decreasing.
To verify (B), consider even and odd terms separately.
$\displaystyle a_{2n-1} = \frac 1 n$, so clearly $\displaystyle \lim_{n\to\infty}a_{2n-1}=0$
$\displaystyle a_{2n}=\int_n^{n+1} \frac {dx} x=\log (\frac{n+1} n)$, so $\displaystyle \lim_{n\to\infty} a_{2n} = 0$.
Therefore, by Leibniz's Rule, the series $\displaystyle \sum(-1)^{(n-1)}a_n$ converges. Denote it's sum by $\displaystyle \gamma$ and it's nth partial sum by $\displaystyle s_n$. Now we inspect the $\displaystyle (2n-1)$st partial sum:
$\displaystyle s_{2n-1}=1-\int_1^2 \frac {dx} x + \frac 1 2 - \int_2^3 \frac {dx} x + \cdots + \frac 1 {n-1} - \int_{n-1}^n \frac {dx} x + \frac 1 n$
Rearranging...
$\displaystyle s_{2n-1}=1 + \frac 1 2 + \cdots + \frac 1 n - \log n$
Since $\displaystyle s_{2n-1} \rightarrow C$ as $\displaystyle n \rightarrow \infty$, we have
$\displaystyle \lim_{n\to\infty}\left(1+\frac 1 2 + \cdots + \frac 1 n -\log n \right)=\gamma$
This $\displaystyle \gamma$ is the Euler-Mascheroni constant. The relation above can also be expressed as
$\displaystyle \sum_{k=1}^n \frac 1 k = \log n + \gamma + o(1)$
This entire post is very similar to the text in Apostol's Calculus, so if there is any concern of copyright infringement (although the book was written in 1969, and up to 10% is reproducible for educational purposes) please feel free to delete this post.