1. ## analysis!!

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

2. Originally Posted by flower3
$\displaystyle { For \ n \in N \ let \ \ x_n = \frac {1}{1+n} + \frac {1}{2+n} +...+\frac {1}{2n} : }$
$\displaystyle a- \ show \ \ that \ \ x_n \ \ is \ \ increasing .$
$\displaystyle b- \ prove \ \ that \ \ x_n < 1 , \ \forall n \in N .$
$\displaystyle c- \ conclude \ \ that \ \ x_n \ \ is \ \ convergent$
Hi

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

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