converge or diverge?

**lindsays** · October 10th 2009, 10:39 AM

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.

**aman_cc** · October 10th 2009, 10:54 AM

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

**lindsays** · October 10th 2009, 11:28 AM

is it possible to prove this using the ratio test?

**Plato** · October 10th 2009, 11:32 AM

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?

**redsoxfan325** · October 10th 2009, 07:51 PM

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.

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