Divergence of a sequence.

Question

In an example in the book, Thomas Calculus 14e:

Q: Show that sequence $\{(-1)^{n+1}\}$ diverges?

A: They choose $\varepsilon$ to be $1/2$ and thus, $|L-1|<1/2$ for $+1$ and $|L+1|<1/2$ if value converges to $+1$, where $L$ is the value to which the sequence converges. On solving both equations, we get no common solution, thus no such limit $L$ exists and thus the sequence diverges.

My question: If we choose $\varepsilon$ to be $3$ or any number greater than it, then the above inequalities give a solution, so does it mean the sequence converges??

Also later it states, a sequence can diverge without diverging to $\pm \infty$, how is that relevant and possible?

Answer 1

No.

The definition of convergence states that it is true for any $\epsilon > 0$, not just for a particular one.

So, if the definition of convergence fails for a particular $\epsilon$, then the sequence diverges.

Answer 2

The definition of convergence of a sequence $\{x_n\}_{n\in\Bbb N}$ to a limit $L$ (say) is that,

For every $\epsilon>0$ there is a $K\in\Bbb N$ such that $$|x_n-L|<\epsilon,\forall n\ge K.$$

Now in the sequence $\{(-1)^{n+1}:n\in\Bbb N\}$ you can not use the term "for every $\epsilon>0$" because there exists an $\epsilon=\frac{1}{2}>0$ and for that $\epsilon $ , there doesnot exists any $K\in\Bbb N$ for which the above relation holds. Hence the sequence diverges (basically oscilates finitely).

Answer 3

A sequence of real numbers $\{a_n\}$ converges to some $a\in\mathbb R$ if given any $\varepsilon\gt 0$, there exists some $N\in\mathbb N $ such that $|a_n-a|\lt \varepsilon$ for all $n\geq N $.

Suppose the sequence $\{(-1)^{n+1}\}$ converges to some $L\in\mathbb R$. So,in particular, given $\varepsilon=\frac 12$, we should be able to find some $N$ such that $|L-(-1)^{n+1}|\lt\frac 12$ for all $n\geq N $. That is, $|L+1|\lt \frac 12$ and $|L-1|\lt \frac 12$. In other words, $\frac {-3}2\lt L\lt \frac {-1}2$ and $\frac 12\lt L\lt \frac {3}2$, which is not possible.

You can do a similar kind of argument with any $\varepsilon$ such that $0\lt \varepsilon \lt 1$. And notice that the sequence does not go to infinity, but oscillates between $1$ and $-1$.