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

[Elzair]

Homework Statement

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

Homework Equations

lim n^2 = a
lim -n = b

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.

 

[jgens]

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

 

[hunt_mat]

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

 

[HallsofIvy]

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

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.

 

[hunt_mat]

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