unbounded sequence

**flower3** · December 9th 2009, 07:39 AM

let $x_n$ be an unbounded sequence . prove that there is a subsequence $x_{n_k}$ of $x_n$ such that $\frac{1}{x_{n_k}} \to 0$

**claves** · December 9th 2009, 07:51 AM

If epsilon > 0 then 1/x_n - 0 < epsilon => x_n > 1/epsilon. If you choose a strictly increasing subsequence of {x_n} (which is possible since it is unbounded), then all but a finite number of x_n > 1/epsilon, so 1/x_n converges to 0.

**Plato** · December 9th 2009, 07:56 AM

Originally Posted by flower3

let $x_n$ be an unbounded sequence . prove that there is a subsequence $x_{n_k}$ of $x_n$ such that $\frac{1}{x_{n_k}} \to 0$

Suppose that $\left(x_n\right)$ is not bounded above.
Then $\left( {\exists x_{N_1 } } \right)\left[ {1 < x_{N_1 } } \right]$ .
Let $B_1 = \max \left\{ {2,N_1 } \right\}$ then $\left( {\exists x_{N_2 } } \right)\left[ {B_1 < x_{N_2 } } \right]$ .
Now you can see how to construct the required subsequence.

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