# converge or diverge?

• Oct 10th 2009, 11:39 AM
lindsays
converge or diverge?
Consider the sequence {a_n} (n is an element of N) given by a_n = (-1)^n. Does the sequence converge or diverge? Prove this.
• Oct 10th 2009, 11:54 AM
aman_cc
Quote:

Originally Posted by lindsays
Consider the sequence {a_n} (n is an element of N) given by a_n = (-1)^n. Does the sequence converge or diverge? Prove this.

Hint - Write down the sequence
-1,1,-1,1,-1,1,-1,1,....

This will go on and on...

So what will you conclude?

To prove it formally, Assume a limit L, prove both a_k and a_k+1 can be <1/2 (or a smaller number) (hence this becomes your epsilon) 'away' from L. Hence a contradiction to the definition of the limit
• Oct 10th 2009, 12:28 PM
lindsays
is it possible to prove this using the ratio test?
• Oct 10th 2009, 12:32 PM
Plato
Quote:

Originally Posted by lindsays
is it possible to prove this using the ratio test?

Just look at the terms of the sequence.
If a sequence converges, the almost all of its terms are very "close together".
If possibility true of this sequence?
• Oct 10th 2009, 08:51 PM
redsoxfan325
Quote:

Originally Posted by lindsays
is it possible to prove this using the ratio test?

The ratio test won't do you any good because it yields $1$, which is inconclusive.

The definition of convergence of a sequence is:

A sequence $\{a_n\}$ converges to a point $a$ iff $\forall~\epsilon>0$, $\exists~M$ such that $m>M$ implies $|a_m-a|<\epsilon$.

Negating this statement gives us:

A sequence $\{a_n\}$ does not converge iff $\exists~\epsilon>0$ such that $\forall~M$, $\exists~ k,m>M$ such that $|a_m-a_k|\geq\epsilon$

Take $\epsilon=1$ and see what happens.