# Math Help - Limits

1. Originally Posted by Plato
OK. Here is the major flaw.
$\color{red} |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M \implies |a_{n+1} - a_n| < M$
There is absolutely no way that you can claim that. It is a false claim.
i can.. take note that $|a_{n+2} - a_{n+1}|$ is always positive, therefore if i remove it, the remaining would be strictly less than M..
where is the flaw there? you are the only person i know who disagreed with that.. as far as i'm concerned..

Originally Posted by Plato
i have explained the reason why that is true..

2. Originally Posted by kalagota
i can.. take note that $|a_{n+2} - a_{n+1}|$ is always positive, therefore if i remove it, the remaining would be strictly less than M..
where is the flaw there? you are the only person i know who disagreed with that.. as far as i'm concerned..

i have explained the reason why that is true..
Thats not the problem. IF this was true
$\color{red} |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M$
we could claim this
$\color{red} \implies |a_{n+1} - a_n| < M$
as you said,
BUT here's the problem:we just know that
$\color{red} |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq M \$
so we can't claim that something greater than $\color{red} |a_{n+2} - a_{n+1} + a_{n+1} - a_n|\$ is less than M, because we don't know that.

3. Originally Posted by Sasbe
[See attached image below.]
I think you need some fairly heavy-duty analysis to deal with this.

Let ε > 0. since $\lim_{n\to\infty}(a_{n+2}-a_n) = 0$, there is an integer N such that $|a_{n+2}-a_n|<\epsilon$ whenever n≥N. For any positive integer r, $a_{N+2r} - a_N = (a_{N+2} - a_N) + (a_{N+4} - a_{N+2}) + (a_{N+6} - a_{N+4}) + \ldots + (a_{N+2r} - a_{N+2r-2}).$

Therefore $|a_{N+2r} - a_N| \leqslant |a_{N+2} - a_N| + |a_{N+4} - a_{N+2}| + \ldots + |a_{N+2r} - a_{N+2r-2}| < r\epsilon.$ It follows that $\frac{|a_{N+2r} - a_N|}{N+2r} < \frac{r\epsilon}{N+2r}\to \epsilon/2$ as r→∞. Hence $\frac{|a_{N+2r} - a_N|}{N+2r} < \epsilon$ for all sufficiently large r. In fact, since $\frac{a_N}{N+2r}\to0$ as r→∞, we even have $\frac{|a_{N+2r}|}{N+2r} < \epsilon$ for all sufficiently large r. Putting n=N+2r (or n=(N+1)+2r, which works equally well), you see that $\frac{|a_n|}n < \epsilon$ for all sufficiently large n.

Finally, $\frac{|a_{n+1} - a_n|}n \leqslant \frac{|a_{n+1}|}n + \frac{|a_n|}n < 2\epsilon$ for all sufficiently large n. Since ε was arbitrary, this shows that $\frac{|a_{n+1} - a_n|}n \to 0$ as n→∞.

4. after re-reading everything, i now realize my big mistake..

thanks to you plato..

Page 2 of 2 First 12