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.

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- Oct 10th 2009, 10:39 AMlindsaysconverge 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, 10:54 AMaman_cc

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, 11:28 AMlindsays
is it possible to prove this using the ratio test?

- Oct 10th 2009, 11:32 AMPlato
- Oct 10th 2009, 07:51 PMredsoxfan325
The ratio test won't do you any good because it yields $\displaystyle 1$, which is inconclusive.

The definition of convergence of a sequence is:

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

Negating this statement gives us:

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

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