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.