1
$\begingroup$

I have to prove that a constant sequence and an eventually cinstant sequence is always convergent.

I tried to do this as follows :

I considered (xn) in X to be a constant sequence such that (xn) = x,x,x.... for all n in N.

Then i considered an open ball centered at x and with radius 1/2 i.e. b[x;1/2).

Since this ball will definitely contain a tail of the sequence (xn) therefore (xn) is convergent.

Is this argument correct.Any arbitrary value of "r" should work, right ?

Also can the convergence of eventually constant sequences be proved on the same lines? How do i decide what value of "r" should be chosen for this case ?

$\endgroup$
2
$\begingroup$

Your argument is not quite correct. You should start with an arbitrary value of $r>0$, at which point your conclusion follows. (Since $r>0$ was arbitrary, then every open ball about $x$ contains a tail of the sequence, so we have convergence.)

As for the eventually constant sequence, you know that there is some $x$ and some $N$ such that $x_n=x$ for all $n\ge N$. The same sort of argument works here, with an arbitrary $r>0$.

To see why we need $r>0$ to be arbitrary, consider the sequence of real numbers $x_n=(-1)^n$. Now, $B[0;2)$ contains every point of the sequence, but $B[0;1)$ contains none of the points. In fact, for every $x\in\Bbb R,$ there is an $R>0$ such that $B[x;R)$ contains every point of the sequence, but there is also some $r>0$ such that $B[x;r)$ does not contain a tail of the sequence, so the sequence does not converge.

$\endgroup$
2
  • $\begingroup$ So is it sufficient to say that the open ball b[x;r) where r is arbitrary will always contain a tail of the sequence (xn) and hence a constant sequence is always convergent. $\endgroup$
    – johny
    Oct 1 '13 at 13:19
  • $\begingroup$ Absolutely fine. $\endgroup$ Oct 1 '13 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.