Unbounded sequence

**tttcomrader** · September 5th 2008, 07:46 AM

Let $x_{1} =1$ , and let $x_{n+1} =\frac {(x_1 +x_2 + . . . + x_{n} ) }{2}$ . Prove that $x_n \rightarrow \infty$

So far:

I have proved that $x_n$ is a strictly decreasing sequence by induction. Will this help a bit?

**Moo** · September 5th 2008, 08:03 AM

Hello,

Originally Posted by tttcomrader

Let $x_{1} =1$ , and let $x_{n+1} =\frac {(x_1 +x_2 + . . . + x_{n} ) }{2}$ . Prove that $x_n \rightarrow \infty$

So far:

I have proved that $x_n$ is a strictly decreasing sequence by induction. Will this help a bit?

Uh... Don't you find it weird that you're asked to prove that $x_n \to +\infty$ and though you find it to be decreasing ?

Consider $n \ge 2$ and see if it's decreasing or increasing ^^

**tttcomrader** · September 5th 2008, 08:40 AM

I have x1=1, x2=1/2, x3=3/4, x4=9/8.

so it is in fact increasing strictly...

**Plato** · September 5th 2008, 08:58 AM

Can you show $k \ge 2\quad \Rightarrow \quad x_k = \frac{{3^{k - 2} }}{{2^{k - 1} }}$ ?

**tttcomrader** · September 6th 2008, 07:59 AM

It is true for k=2.

If it is true for k=n, then we have $x_k = \frac {x_1 + . . . + x_{k-1} }{2}= \frac {3^{k-2}}{2^{k-1}}$

Then $x_{k+1} = \frac {x_1 + . . . + x_{k-1} + x_{k} } {2} = \frac {3 ^ {k-2} }{2^{k-1}} + \frac {x_k}{2} = \frac {3 ^ {k-2} }{2^{k-1}} + \frac {3^{k-2}}{2^{k}}$

$= \frac { (2)3^{k-2} +3^{k-2} } {2^k} = \frac {3^ {k-1}}{2^k} = \frac {3^{(k+1)-2}} {2^{(k+1)-1}}$

So it is true for all $k \in \mathbb {N}$

Here is my proof to finish the problem:

$x_n = \frac {3^{k-2}}{2^{k-1}} = \frac {2}{3^{2}} ( \frac {3}{2})^k \geq \frac {2}{3^{2}} \ \ \ \ \ \ \forall k \in \mathbb {N}$

So this sequence is not bounded.

Is this correct? Thank you.

Thread: Unbounded sequence

LinkBack

Thread Tools

Display

Unbounded sequence

Similar Math Help Forum Discussions

[SOLVED] to prove that the sequence is unbounded.

unbounded sequence

Bounded sequence whose set of acc points is unbounded.

Unbounded Sequence

Unbounded sequence?

Search Tags