• Nov 11th 2009, 06:39 AM
chisigma
Given the Euler's constant...

$\displaystyle \gamma = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n} \frac{1}{k} - \ln n \}= .577215664901...$ (1)

... demonstrate that is...

$\displaystyle \gamma = \sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k)}{k}$ (2)

... where...

$\displaystyle \zeta(k)= \sum_{n=1}^{\infty} \frac{1}{n^{k}}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Nov 11th 2009, 07:02 AM
NonCommAlg
Quote:

Originally Posted by chisigma
Given the Euler's constant...

$\displaystyle \gamma = \lim_{n \rightarrow \infty} \{\sum_{k=1}^{n} \frac{1}{k} - \ln n \}= .577215664901...$ (1)

... demonstrate that is...

$\displaystyle \gamma = \sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k)}{k}$ (2)

... where...

$\displaystyle \zeta(k)= \sum_{n=1}^{\infty} \frac{1}{n^{k}}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

for any $\displaystyle -1 < x \leq 1$ we have $\displaystyle \sum_{k=2}^{\infty} \frac{(-x)^k}{k}=x - \ln(1+x),$ which is proved by integrating $\displaystyle \frac{1}{1+x}=1-x + x^2 - \cdots .$

putting $\displaystyle x=\frac{1}{n}, \ n \in \mathbb{N},$ gives us $\displaystyle \sum_{k=2}^{\infty} \frac{(-1)^k}{kn^k} = \frac{1}{n} - \ln \left(1 + \frac{1}{n} \right).$ therefore:

$\displaystyle \sum_{k=2}^{\infty} (-1)^k \frac{\zeta(k)}{k}=\sum_{k=2}^{\infty} \frac{(-1)^k}{k} \sum_{n=1}^{\infty} \frac{1}{n^k}= \sum_{n=1}^{\infty} \sum_{k=2}^{\infty} \frac{(-1)^k}{kn^k}$

$\displaystyle =\sum_{n=1}^{\infty} \left [ \frac{1}{n} - \ln \left(1 + \frac{1}{n} \right) \right]=\lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{k} - \ln(n+1) \right] = \gamma.$