# Proving {x^2} & {-n} do not converge.

1. Sep 9, 2010

### Elzair

1. The problem statement, all variables and given/known data

Prove the sequences $$\left\{ n^{2} \right} _{n \in N}$$ and $$\left{ -n \right}_{n \in N}$$ do not converge.

2. Relevant equations

lim n^2 = a
lim -n = b

3. The attempt at a solution

If n^2 converges, then for all epsilon > 0, there is an N s.t. |n^2 - a| < epsilon for all n > N.

If -n converges, then for all epsilon > 0, there is an N s.t. |-n - b| < epsilon for all n > N.

I assume I could use the triangle inequality possibly.

2. Sep 9, 2010

### jgens

Why not just pick some concrete value for $\varepsilon$ (say, 1) and derive a contradiction?

3. Sep 9, 2010

### hunt_mat

If a sequence diverges then there is an M>0 thuch that when $$n>N$$ then $$|a_{n}|\geqslant M$$. Can you do this for the sequences you have?

4. Sep 9, 2010

### HallsofIvy

Staff Emeritus
Be carful. In general a sequence "diverges" as long as it does not converge. That is very different from "diverges to infinity" or "diverges to negative infinity". The sequence 1, 2, 1, 2, 1, 2, ..., diverges but does NOT satify the condition you give.

5. Sep 10, 2010

### hunt_mat

But in the context here the definition of diverges to infinity is the correct one to use.