# Math Help - Sequence Convergence

1. ## Sequence Convergence

Hello all,

How do I show that if the distances between the terms of a sequence are getting smaller, then the sequence is convergent?
In other words, if $|a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$ for all n, then $a_n$ is a convergent sequence.
In fact, I have the following inequality: $\frac{|a_{n+1}-a_{n}|}{4}\leq |a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$, and I know that $1\leq a_n\leq 2$

Thank you

Hello all,

How do I show that if the distances between the terms of a sequence are getting smaller, then the sequence is convergent?
In other words, if $|a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$ for all n, then $a_n$ is a convergent sequence.
In fact, I have the following inequality: $\frac{|a_{n+1}-a_{n}|}{4}\leq |a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$, and I know that $1\leq a_n\leq 2$

Thank you

More background would be nice. If you can prove it's Cauchy you may conclude convergence since $\mathbb{R}$ is complete.

3. For the sequence $(a_n)=(-1)^n$, we have $|a_{n+2}-a_{n+1}|=|a_{n+1}-a_n|$ (and thus $\leq$)
But the sequence is not convergent :P

Hello all,

How do I show that if the distances between the terms of a sequence are getting smaller, then the sequence is convergent?
In other words, if $|a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$ for all n, then $a_n$ is a convergent sequence.
In fact, I have the following inequality: $\frac{|a_{n+1}-a_{n}|}{4}\leq |a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$, and I know that $1\leq a_n\leq 2$

Thank you

Doesn't the sequence $a_n=\sum_{k=0}^n\frac1{k}$ fit your hypothesis? Does it converge?

You need something a little stronger to conclude that a sequence is convergent. For example, if $|a_{n+2}-a_{n+1}|\leq\alpha|a_{n+1}-a_n|$ for some $\alpha<1$, then the sequence is Cauchy.

5. I'm sorry guys. It seems like I the things I omitted were important.
Well, the problem says: $a_n$ is a sequence defined by: $a_1 = 1$ and $a_{n+1} = 1+\frac{1}{a_n}$. Find an inequality between $|a_{n+2}-a_{n+1}|$ and $|a_{n+1}-a_{n}|$, then use the inequality to prove that the sequence is $a_n$ is convergent.
First, we can note that $1\leq a_n \leq 2$ for all n. From this, we can find the following inequalities: $\frac{|a_{n+1}-a_{n}|}{4} \leq |a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$.
I tried to prove that the sequence is Cauchy, but the the bound I'm getting on $|a_n - a_m|$ depends on n and m.

Thank you..

I'm sorry guys. It seems like I the things I omitted were important.
Well, the problem says: $a_n$ is a sequence defined by: $a_1 = 1$ and $a_{n+1} = 1+\frac{1}{a_n}$. Find an inequality between $|a_{n+2}-a_{n+1}|$ and $|a_{n+1}-a_{n}|$, then use the inequality to prove that the sequence is $a_n$ is convergent.
First, we can note that $1\leq a_n \leq 2$ for all n. From this, we can find the following inequalities: $\frac{|a_{n+1}-a_{n}|}{4} \leq |a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$.
I tried to prove that the sequence is Cauchy, but the the bound I'm getting on $|a_n - a_m|$ depends on n and m.
That makes a big difference.
Use induction to show that the sequence increasing and bounded above by 2.
That will show that it converges.
Solve $L=1+\frac{1}{L}$ to find the limit.

7. Hello Plato,

But the problem is that the sequence is not monotone. That's why we are asked to find the inequality between the differences. We notice that these differences are getting smaller and smaller. How can we conclude that the sequence is convergent?

Thank you,

Hello Plato,

But the problem is that the sequence is not monotone. That's why we are asked to find the inequality between the differences. We notice that these differences are getting smaller and smaller. How can we conclude that the sequence is convergent?

Thank you,
You cannot conclude that a sequence is convergent simply because the differences between successive terms is getting smaller. For instance, define a sequence recursively by $a_1=1,\:a_{n+1}=a_n+1+\frac1n$. Clearly the difference between successive terms is strictly decreasing, but it's always greater than 1, so obviously this sequence cannot converge.

But even if the limit of the difference between successive terms goes to zero, you still cannot conclude convergence; the harmonic series is the standard example.

9. I'm sorry guys. I really need help with this one. I will repeat the problem: $a_n$ is a sequence defined by: $a_1 = 1$ and $a_{n+1} = 1+\frac{1}{a_n}$. Find an inequality between $|a_{n+2}-a_{n+1}|$ and $|a_{n+1}-a_{n}|$, then use the inequality to prove that the sequence is $a_n$ is convergent.
First, we can note that $1\leq a_n \leq 2$ for all n. From this, we can find the following inequalities: $\frac{|a_{n+1}-a_{n}|}{4} \leq |a_{n+2}-a_{n+1}|\leq |a_{n+1}-a_{n}|$.
I tried to prove that the sequence is Cauchy, but the the bound I'm getting on $|a_n - a_m|$ depends on n and m.
Note that the sequence is not monotone, that's why we were asked to find above inequality.

Thank you..

10. For instance,

$a_{n+2}-a_{n+1}=\left(1+\frac{1}{a_{n+1}}\right)-\left(1+\frac{1}{a_{n}}\right)$ $=\frac{a_n-a_{n+1}}{a_na_{n+1}}$,

and $a_n a_{n+1}=a_n+1$ by the recurrence equation, hence if you know $1\leq a_n\leq 2$, then $2\leq a_na_{n+1}\leq 3$.

From this, you can conclude.

Sidenote: $a_n=1+\frac{1}{a_{n-1}}=1+\cfrac{1}{1+\cfrac{1}{a_{n-2}}}=\cdots=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+ \cdots+\cfrac{1}{1}}}}$ with $n-1$ fractions.

11. Hey all
I understood the idea that having $|a_{n+2} - a_{n+1}| \leq |a_{n+1} - a_{n}|$ for all n is not enough for convergence. What if we have: $|a_{n+2} - a_{n+1}| \leq \frac{|a_{n+1} - a_{n}|}{2}$?

Thanx

What if we have: $|a_{n+2} - a_{n+1}| \leq \frac{|a_{n+1} - a_{n}|}{2}$?
Use this equation recursively to bound $|a_{n+2}-a_{n+1}|$ by a geometric sequence, and then to bound $|a_{n+m}-a_n|\leq |a_{n+m}-a_{n+m-1}|+\cdots+|a_{n+1}-a_n|$ by a sequence converging to 0 as $n\to\infty$, to prove that $(a_n)_n$ is a Cauchy sequence.