If {x_n} is not bounded above, does x_n -> infinity? prove there exist a subsequence {x_n} where k=1 to infinity such that x_n_k -> infinity
To determine whether the sequence {x_n} converges to infinity, we first need to understand the concept of boundedness. A sequence is said to be bounded above if there exists a real number M such that all terms of the sequence are less than or equal to M. If a sequence is not bounded above, it means that there is no such real number M that can serve as an upper bound for all its terms.
Now, if the sequence {x_n} is not bounded above, it does not necessarily imply that the sequence diverges to infinity. It could still converge to a finite value, diverge to negative infinity, oscillate, or exhibit other divergent behavior.
To prove the existence of a subsequence {x_n_k} where k ranges from 1 to infinity that converges to infinity, we can use the fact that {x_n} is not bounded above. Since there is no upper bound, we can construct a subsequence that keeps selecting larger terms from the original sequence.
Here's how we can do it:
1. Start with k = 1.
2. Since {x_n} is not bounded above, we can select the first term, x_1.
3. Now, consider the subsequence {x_n_k} where k = 1.
4. Select the term x_n_1 such that x_n_1 > x_1.
5. Increment k to 2.
6. Choose the term x_n_2 such that x_n_2 > x_n_1.
7. Continue this process recursively, selecting terms x_n_k such that x_n_k > x_n_k-1 for each k.
8. Since {x_n} is not bounded above, there will always be a term that is greater than the previous one.
9. By construction, the subsequence {x_n_k} will be an increasing sequence with no upper bound.
10. Therefore, as k approaches infinity, x_n_k will also approach infinity.
Hence, we have found a subsequence {x_n_k} where k ranges from 1 to infinity that converges to infinity, given that the original sequence {x_n} is not bounded above.