# [SOLVED] Convertence

• Nov 3rd 2007, 07:01 PM
putnam120
[SOLVED] Convertence
Let $\displaystyle a_n$ be a sequence of real numbers such that $\displaystyle |a_{n+1}-a_n|<\frac{1}{2^n}$ for $\displaystyle n=1,2,\dots$ prove that the sequence converges.

My first thought was to show that $\displaystyle a_n$ was a Cauchy sequence but am having trouble doing so. Any ideas about what else I should try?
• Nov 3rd 2007, 07:15 PM
Jhevon
Quote:

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

My first thought was to show that $\displaystyle a_n$ 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 $\displaystyle |a_n - a_m| = |a_n - a_{n + 1} + a_{n + 1} - a_{n + 2} + a_{n + 2} - \cdots + a_{m - 1} - a_m|$

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

can you continue?
• Nov 3rd 2007, 07:29 PM
ThePerfectHacker
Quote:

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

My first thought was to show that $\displaystyle a_n$ 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 $\displaystyle a_n$ be a sequence so that:
$\displaystyle \lim \ |a_{n+1}-a_n| = \frac{1}{2}$
Prove that $\displaystyle a_n$ is a convergent sequence.

It is extremely similar to your problem. What Jhevon said is basically the solution, you should use Cauchy sequence.
• Nov 3rd 2007, 07:33 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
In my Real Analysis course we had the following problem on the exam:
Let $\displaystyle a_n$ be a sequence so that:
$\displaystyle \lim \ |a_{n+1}-a_n| = \frac{1}{2}$
Prove that $\displaystyle a_n$ 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
• Nov 3rd 2007, 07:59 PM
putnam120
So from the hint I have that
$\displaystyle |a_n-a_m|\le |a_n-a_{n+1}+\dots -a_m|\le |a_n-a_{n+1}|+\dots + |a_{m-1}-a_m| < \frac{1}{2^n}+\dots +\frac{1}{2^{m-1}} <$ $\displaystyle \frac{1}{2^{n-1}}$ for $\displaystyle n>m$.

Is that right?
• Nov 3rd 2007, 08:05 PM
ThePerfectHacker
Quote:

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

Is that right?

No.

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

Now follow by the geometric sum.
• Nov 3rd 2007, 08:33 PM
putnam120
ok i was just being lazy. but thanks 4 all the help. :D