1. ## [SOLVED] Convertence

Let $a_n$ be a sequence of real numbers such that $|a_{n+1}-a_n|<\frac{1}{2^n}$ for $n=1,2,\dots$ prove that the sequence converges.

My first thought was to show that $a_n$ was a Cauchy sequence but am having trouble doing so. Any ideas about what else I should try?

2. Originally Posted by putnam120
Let $a_n$ be a sequence of real numbers such that $|a_{n+1}-a_n|<\frac{1}{2^n}$ for $n=1,2,\dots$ prove that the sequence converges.

My first thought was to show that $a_n$ was a Cauchy sequence but am having trouble doing so. Any ideas about what else I should try?
that is what you should try. here, i'll start you off.

Note that $|a_n - a_m| = |a_n - a_{n + 1} + a_{n + 1} - a_{n + 2} + a_{n + 2} - \cdots + a_{m - 1} - a_m|$

........................... $\le |a_n - a_{n + 1}| + |a_{n + 1} - a_{n + 2}| + \cdots + |a_{m - 1} - a_m|$ by the Triangle inequality

can you continue?

3. Originally Posted by maokid
Let $a_n$ be a sequence of real numbers such that $|a_{n+1}-a_n|<\frac{1}{2^n}$ for $n=1,2,\dots$ prove that the sequence converges.

My first thought was to show that $a_n$ was a Cauchy sequence but am having trouble doing so. Any ideas about what else I should try?
In my Real Analysis course we had the following problem on the exam:
Let $a_n$ be a sequence so that:
$\lim \ |a_{n+1}-a_n| = \frac{1}{2}$
Prove that $a_n$ is a convergent sequence.

It is extremely similar to your problem. What Jhevon said is basically the solution, you should use Cauchy sequence.

4. Originally Posted by ThePerfectHacker
In my Real Analysis course we had the following problem on the exam:
Let $a_n$ be a sequence so that:
$\lim \ |a_{n+1}-a_n| = \frac{1}{2}$
Prove that $a_n$ is a convergent sequence.

It is extremely similar to your problem. What Jhevon said is basically the solution, you should use Cauchy sequence.
yes, i remember that exam question, i liked the solution, for some reason, that's why i remembered it.

@putnam120: another hint: note that you will need to use the sum formula for a geometric series here, so think about that if you get stuck

5. So from the hint I have that
$|a_n-a_m|\le |a_n-a_{n+1}+\dots -a_m|\le |a_n-a_{n+1}|+\dots + |a_{m-1}-a_m| < \frac{1}{2^n}+\dots +\frac{1}{2^{m-1}} <$ $\frac{1}{2^{n-1}}$ for $n>m$.

Is that right?

6. Originally Posted by putnam120
So from the hint I have that
$|a_n-a_m|\le |a_n-a_{n+1}+\dots -a_m|\le |a_n-a_{n+1}|+\dots + |a_{m-1}-a_m| < \frac{1}{2^n}+\dots +\frac{1}{2^{m-1}} <$ $\frac{1}{2^{n-1}}$ for $n>m$.

Is that right?
No.

$|a_{n+1}-a_n| + ... + |a_{m+1}-a_m| \leq \frac{1}{2^n}+...+\frac{1}{2^m} = \frac{1}{2^m}\left( 1 + \frac{1}{2}+...+\frac{1}{2^{n-m}} \right)$.

Now follow by the geometric sum.

7. ok i was just being lazy. but thanks 4 all the help.