I really need help with this one. Is there anyone who can get me started? thanks
Well, it's been a few weeks since I've done limits, but if $\displaystyle n$ goes to infinite, then you're basically having every $\displaystyle n$ variable go off to 0, so if you substitute 0 in, you're left with $\displaystyle lim a(0)+1-a(0)/0=0$ and $\displaystyle 1/0=0$ so that should be your proof.
i'll try to prove it like this.. since $\displaystyle \lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0$, then $\displaystyle a_{n+2} - a_n$ is bounded, say by M
$\displaystyle \implies |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M \implies |a_{n+1} - a_n| < M $ (why?) $\displaystyle \implies -M < a_{n+1} - a_n < M $
so when we multiply $\displaystyle \frac{1}{n}$ to all, we have $\displaystyle \frac{-M}{n} < \frac{a_{n+1} - a_n}{n} < \frac{M}{n} $ and if $\displaystyle n \rightarrow \infty$, we have, $\displaystyle 0 < \frac{a_{n+2} - a_n}{n} < 0 $ and by squeeze theorem, we get $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0 $. QED
It may indeed matter. Take the example $\displaystyle a_n = \left( { - 1} \right)^n + \frac{1}{n}$.
It is clear that both n & n+2 are together even or odd.
So $\displaystyle \left( {a_{n + 2} - a_n } \right) \to 0$.
However, $\displaystyle \left( {a_{n + 1} - a_n } \right)$ does not converge to zero.
If n is even we get $\displaystyle \left( {a_{n + 1} - a_n } \right) = \left[ { - 1 + \frac{1}{{n + 1}}} \right] - \left[ {1 + \frac{1}{n}} \right] = - 2 + \frac{1}{{n + 1}} - \frac{1}{n}$.
Whereas if n is odd we get $\displaystyle \left( {a_{n + 1} - a_n } \right) = \left[ {1 + \frac{1}{{n + 1}}} \right] - \left[ { - 1 + \frac{1}{n}} \right] = 2 + \frac{1}{{n + 1}} - \frac{1}{n}$.
The key here is being bounded. That is what needs to be shown for $\displaystyle \left( {a_n } \right)$
How do we know this?
$\displaystyle |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M$
I would understand this if $\displaystyle |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n|$ was less (not greater) than $\displaystyle |a_{n+2} - a_{n+1} + a_{n+1} - a_n|$, but its not.
The whole point is that $\displaystyle \left( {a_n } \right)$ may not be a covergent sequence.
Take a close look at the example that I gave above.
In that example the $\displaystyle \left( {a_n } \right)$ sequence does not converge. Now it is bounded.
So from the given, you may not conclude that the sequence converges because there is a counterexample.
How do you prove that it is bounded?
Theorem: A convergent sequence is bounded.. (do i need to prove this?)
the given was $\displaystyle \lim_{n \rightarrow \infty} a_{n+2} - a_n $ and not $\displaystyle {a_n } $ alone..
you should not put your interest with $\displaystyle a_n$ since what we want is the sequence $\displaystyle a_{n+1} - a_n$ and that sequence is different from $\displaystyle a_n$!
if you find any "flaws" with my "proof" while i had used only the given and the definition and some theorems, point it out.. well, i would try to consult this with my professor tomorrow..
OK. Here is the major flaw.
$\displaystyle \color{red} |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M \implies |a_{n+1} - a_n| < M $
There is absolutely no way that you can claim that. It is a false claim.
Don't you see? Sasbe had already pointed that out to you.