# limit, sequence, exists

• Apr 15th 2009, 07:49 PM
zelda2139
limit, sequence, exists
Define $a_n = 1+\frac{1}{2}+\frac{1}{3}+ \cdots + \frac{1}{n} - \int^n_1 \frac{1}{t}dt$. Prove that $a=\lim_{n \rightarrow \infty} a_n$ exists and $0.
Attempt
My issue is the integral in the sequence. I don't know how to take the limit of the sequence with that in there. I think there is a way to break up the integral to take the limit of the sequence easier. I need help with this question. Thanks in advance.
• Apr 16th 2009, 12:01 AM
chisigma
Let's indicate with $\gamma$ the limit...

$\gamma = \lim_{n\rightarrow \infty} 1 + \frac {1}{2} + \frac{1}{3} + \dots + \frac{1}{n} - \int_{1}^{n}\frac{dt}{t} = \lim_{n\rightarrow \infty} \int_{1}^{n} (\frac {1}{\lfloor t \rfloor} - \frac{1}{t})\cdot dt = \lim_{n\rightarrow \infty} \sum_{k=1}^{n} \int_{k}^{k+1} (\frac {1}{\lfloor t \rfloor} - \frac{1}{t})\cdot dt$

... where $\lfloor * \rfloor$ is the so called 'floor function'. On the basis of geometrical considerations [a diagram would be useful to better undestand...] you can demonstrate that is...

$\int_{k}^{k+1} (\frac {1}{\lfloor t \rfloor} - \frac{1}{t})\cdot dt < \frac {1}{2\cdot k^{2}}$

... so that...

$\gamma < \lim_{n\rightarrow \infty} \frac{1}{2}\cdot \sum_{k=1}^{n} \frac{1}{k^{2}}= \frac{\pi^{2}}{12} = .822467033424 \dots$

Kind regards

$\chi$ $\sigma$