2
$\begingroup$

My goal is to prove the monotone convergence of a non-increasing sequence of real numbers. There are some steps in the proof that I'm not sure about. The question:

If $S$ is a non-increasing sequence bounded below, show that it converges.

Here is what I have so far:

Since S is non-increasing and bounded below, by the Greatest Lower Bound property, it has an inf.

Let $c$ = inf($S$). Goal: Show that $S$ converges to $c$.

By definition of infimum, for every $\epsilon$ $> 0$, $\exists N$ such that $S_N$ $< c + \epsilon$ (Otherwise, $c + \epsilon$ would be a lower bound of $S$ and that contradicts $c$ being the infimum of $S$.

Here is where I am unsure about my proof:

Because $S$ is non-increasing, if $n > N$, $\epsilon > \mid S_n - c \mid \geq \mid S_N - c\mid$.

So, that proves that $S$ converges to $c$, which is inf($S$).

$\endgroup$
1
$\begingroup$

You are correct for the most part. You just have to be careful about the final inequalities -- switch $n$ and $N$.

Since the sequence is non-increasing, if $n \geq N$ then $S_n \leq S_N$, and

$c \leq S_n < c + \epsilon$.

So

$0 \leq S_n-c < \epsilon$

and

$|S_n - c| < \epsilon$.

$\endgroup$
  • $\begingroup$ Thanks. Now that I think about it, it makes sense! $\endgroup$ – user3025403 May 10 '14 at 21:26

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.