analysis!!

**flower3** · September 10th 2009, 11:27 AM

${ For \ n \in N \ let \ \ x_n = \frac {1}{1+n} + \frac {1}{2+n} +...+\frac {1}{2n} : }$
$a- \ show \ \ that \ \ x_n \ \ is \ \ increasing .$
$b- \ prove \ \ that \ \ x_n < 1 , \ \forall n \in N .$
$c- \ conclude \ \ that \ \ x_n \ \ is \ \ convergent$

**running-gag** · September 10th 2009, 11:42 AM

Originally Posted by flower3

${ For \ n \in N \ let \ \ x_n = \frac {1}{1+n} + \frac {1}{2+n} +...+\frac {1}{2n} : }$
$a- \ show \ \ that \ \ x_n \ \ is \ \ increasing .$
$b- \ prove \ \ that \ \ x_n < 1 , \ \forall n \in N .$
$c- \ conclude \ \ that \ \ x_n \ \ is \ \ convergent$

Hi

a- Compute $x_{n+1}-x_{n}$
b- $x_n$ is the sum of n term, each of them being lower than ...
c- Use the appropriate theorem

**Veve** · September 10th 2009, 11:49 AM

a) just compute $x_{n+1}-x_n$ and conclude that $\ge0$
b) $\frac{1}{n+1}\le\frac{1}{n+1}, \frac{1}{n+2}<\frac{1}{n+1}, ...,\frac{1}{2n}<\frac{1}{n+1}$ after we sum them we get $x_n<\frac{n}{n+1}<1$
c) we have an increasing sequence $(x_n)_n$ and bounded by 0 and 1, so $(x_n)_n$ is convergent.

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