Convergence of a Monotonic Sequence: Proof and Example

[spenghali]

Homework Statement

If a_{1} = 1 and a_{n+1} = (1-(1/2^{n})) a_{n}, prove that a_{n} converges.

Homework Equations

NONE

The Attempt at a Solution

I am confident about my attempt, I just want it checked. Thanks.

First show that a_{n} is monotone: a_{n} = {1, 1/4, 21/32, 315/512,...}

Claim: a_{n+1} \leq a_{n} for all n \in N. (Must show this)

Proof:

a_{n+1} \leq a_{n}

\Leftrightarrow (1-(1/2^{n})) a_{n} \leq a_{n}

\Leftrightarrow a_{n} - (a_{n} / 2^{n}) \leq a_{n}

\Leftrightarrow 2^{n} a_{n} - a_{n} \leq 2^{n} a_{n}

\Leftrightarrow 2^{n} a_{n} - 2^{n} a_{n} \leq a_{n}

\Leftrightarrow 0 \leq a_{n}

So we can conclude that a_{n+1} \leq a_{n} as long as a_{n} \geq 0. Now a_{1} = 1, so the sequence starts with a number greater than or equal to zero. Every other number in the sequence has the form:

x(1-(1/2^{x})) for x greater than zero.

We claim every such number is greater than or equal to zero, and will show this by induction.

Induction Step: Assume a_{n} \geq 0

a_{n+1} = a_{n} (1-(1/2^{n})) \geq 0(1-(1/2)) = 0

This implies a_{n+1} \geq a_{n}, so since a_{n} \geq 0 for all n \in N, this implies that a_{n} is bounded and monotonic. So by the Monotone Sequence Theorem, a_{n} converges for all n \in N ■

 

[Hurkyl]

Your proof looks reasonable.

You misstated the conclusion though -- For each n, a_n is a number so it doesn't make sense to say "a_n converges for all n". What you meant is that the sequence a_n converges.

 

[spenghali]

ah yes, thanks for the tip.