# Thread: unbounded sequence

1. ## unbounded sequence

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

2. 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.

3. Originally Posted by flower3
let $\displaystyle x_n$ be an unbounded sequence . prove that there is a subsequence $\displaystyle x_{n_k}$ of $\displaystyle x_n$ such that $\displaystyle \frac{1}{x_{n_k}} \to 0$
Suppose that $\displaystyle \left(x_n\right)$ is not bounded above.
Then $\displaystyle \left( {\exists x_{N_1 } } \right)\left[ {1 < x_{N_1 } } \right]$.
Let $\displaystyle B_1 = \max \left\{ {2,N_1 } \right\}$ then $\displaystyle \left( {\exists x_{N_2 } } \right)\left[ {B_1 < x_{N_2 } } \right]$.
Now you can see how to construct the required subsequence.