# Convergence of a sequence

1. Dec 9, 2012

### Felafel

I think the solution I've found makes sense, but I'd like it to be double-checked.
1. The problem statement, all variables and given/known data

Let $(a_n)$ be a limited sequence and $(b_n)$ such that $0≤b_n≤ \frac{1}{2} B_{n-1}$

Prove that if
$a_{n+1} \ge a_{n} -b_{n}$

Then
$\lim_{n\to \infty}a_n$

exists.

3. The attempt at a solution

I can say that $b_n \ge (\frac{1}{2})^n b_0$ which is constant.
THen,
$a_n-a_{n+1} \ge (\frac{1}{2})^nb_0$

Thus $|a_n-a{n+1}|$ is a Cauchy sequence, which means it converges and therefore the limit exists.

2. Dec 9, 2012

### Michael Redei

You need to watch your inequality signs a bit. From $0\leq b_n\leq\frac12b_{n-1}$ you get $0\leq b_n\leq(\frac12)^nb_0$. And that isn't a constant, since it depends on $n$. What you probably mean is that $b_0$ is constant.

Then, $a_n-a_{n+1}\leq(\frac12)^nb_0$. But how do you now conclude that $(|a_n-a_{n+1}|)_{n\in\mathbb N}$ is a Cauchy sequence? And even if it is, how does that make the sequence $(a_n)_{n\in\mathbb N}$ converge?

3. Dec 9, 2012

### Felafel

Right, $\frac{1}{2}^nb_0$ is not a constant, but for n diverging to $\infty b_0$ gets smaller and smaller, so $a_n-a_{n+1}$ is bounded above.
For Cauchy's criterion a sequence converges to something if and only if this holds:
for every  $\epsilon> 0$ there exists n* such that $|a_n-a_m| < \epsilon$ whenever n, m > n*.
Let's just call $\frac{1}{2}b_0=\epsilon$ and $a_m=a_{n+1}$. for a n*, big enough, the criterion holds.

4. Dec 9, 2012

### pasmith

That's not sufficient.

You need to show that for all $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n,m \geq N$, $|a_n - a_m| < \epsilon$. You can't just choose a particular $\epsilon$.

Hint: Without loss of generality you can take $m > n$. Then
$$|a_n - a_m| = |a_n - a_{n+1} + a_{n+1} + \dots - a_{m-1} + a_{m-1} - a_m|$$

Last edited: Dec 9, 2012
5. Dec 9, 2012

### Felafel

ok, what if I prove it this way:
by definition, a sequent is convergent when the following property holds:
for every $\epsilon>0$ there exists an $N$ such that for every n$\geq N$, we have
$|a_n-L|<\epsilon$.
In this exercise I have $|a_n - a_m| < \epsilon$
but by taking
$n,m \geq N$
I can obtain
$|a_n-a_m+L-L|<\epsilon$which is $\geq |a_n-L|+|a-M-L| \geq \frac{\epsilon}{2}+\frac{\epsilon}{2}$
is this what you meant?

6. Dec 9, 2012

### Michael Redei

$b_0$ doesn't get smaller, ever. You really should be more careful about what you put in words, because you need even more care with formulas.

You don't have $|a_n-a_m|<\epsilon$. That's what you want to get. What you're using here is called "circular reasoning", and it's a very easy mistake to make -- confusing what you are aiming at with what you're starting off with.

I think you might try pasmith's suggestion: Assume $m>n$ and have a look at
$$|a_n-a_m| = |a_n-a_{n-1}+a_{n-1}-a_{n-2}+a_{n-2}-\ldots-a_{m+2}+a_{m+2}-a_{m+1}+a_{m+1}-a_m|$$

7. Dec 11, 2012

### Felafel

thanks! is it okay now?:
$|a_n - a_m| \le |a_n - a_{n-1}| + |a_{n-1}-a_{n-2}|+...+|a_m - a_{m-1}|$
each member of this sum is ≤ $(\frac{1}{2})^nb_0$ thus $|a_n-a_m|$ converges.
(i don't know exactly how to formally find $\epsilon$
can I also add that $b_n$ goes to 0 due to the squeeze rule?
thanks again

8. Dec 11, 2012

### Michael Redei

You must check your inequalities better. Each member of that sum is $\leq(\frac12)^{n-1}b_0$ (note the exponent).

Also, you don't find $\epsilon$. That value is just assumed to be positive, and you must show that the sum is smaller than $\epsilon$. So you need to ensure that
$$\left(\frac12\right)^{n-1}b_0 < \epsilon.$$
Suppose you solve that and get some integer $N$ as a result. Then you can re-write your proof "backwards":

Let $\epsilon>0$ be given. Then we set $N=(\ldots \epsilon \ldots)$, and for $n,m\geq N$ with $m>n$ we have
$$\epsilon > \left(\frac12\right)^Nb_0 \geq \ldots \geq |a_n-a_m|.$$
This proves that $(a_n)$ is a Cauchy sequence, and so this sequence must converge. QED

9. Dec 11, 2012

### Felafel

thank you, i really didn't get it