Euler's constant.

**Also sprach Zarathustra** · May 23rd 2011, 04:48 AM

Hello!

How to prove that $a_n=\Sigma^n_{k=1}\frac{1}{k}-log(n)$
is a bounded sequence. ( $0 \leqslant a_n \leqslant 1$ )

Thanks!

**chisigma** · May 23rd 2011, 05:36 AM

Originally Posted by Also sprach Zarathustra

Hello!

How to prove that $a_n=\Sigma^n_{k=1}\frac{1}{k}-log(n)$
is a bounded sequence. ( $0 \leqslant a_n \leqslant 1$ )

Thanks!

Taking into account the basic identity...

$\gamma= \lim_{k \rightarrow \infty} \sum_{n=1}^{k} (\frac{1}{n} - \ln n) = \int_{1}^{\infty} (\frac{1}{\lfloor\ x\ \rfloor} - \frac{1}{x})\ dx$ (1)

... solving the integral in (1) You obtain...

$\gamma= 1 + \sum_{n=2}^{\infty} (\frac{1}{n} - \ln \frac{n}{n-1})$ (2)

Now because for $n \ge 2$ is...

$\ln \frac{n}{n-1} = \frac{1}{n-1} - \frac{1}{2\ (n-1)^{2}} + ... > \frac{1}{n}$ (3)

... each term of the 'infinite sum' in (2) is <0 and therefore is $\gamma<1$ ...

Kind regards

$\chi$ $\sigma$

**FernandoRevilla** · May 23rd 2011, 05:41 AM

Hint Compare $\sum_{n=1}^N1/n$ with $\int_1^Ndx/x=\log N$ , you'll obtain

$\log (N+1)\leq \sum_{n=1}^N1/n\leq 1+\log N$

Edited: Sorry, I didn't see chisigma's post.

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