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## Homework Statement

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

## Homework Equations

NONE

## The Attempt at a Solution

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

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

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

Proof:

[tex]a_{n+1}[/tex] [tex]\leq[/tex] [tex]a_{n}[/tex]

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

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

[tex]\Leftrightarrow[/tex] [tex]2^{n}[/tex] [tex]a_{n}[/tex] - [tex]a_{n}[/tex] [tex]\leq[/tex] [tex]2^{n}[/tex] [tex]a_{n}[/tex]

[tex]\Leftrightarrow[/tex] [tex]2^{n}[/tex] [tex]a_{n}[/tex] - [tex]2^{n}[/tex] [tex]a_{n}[/tex] [tex]\leq[/tex] [tex]a_{n}[/tex]

[tex]\Leftrightarrow[/tex] 0 [tex]\leq[/tex] [tex]a_{n}[/tex]

So we can conclude that [tex]a_{n+1}[/tex] [tex]\leq[/tex] [tex]a_{n}[/tex] as long as [tex]a_{n}[/tex] [tex]\geq[/tex] 0. Now [tex]a_{1}[/tex] = 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/[tex]2^{x}[/tex])) 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 [tex]a_{n}[/tex] [tex]\geq[/tex] 0

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

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