Proving Induction for 1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n}}

[Gooolati]

Homework Statement

Prove By Induction that for all n:

1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n}} = 2 - \frac{1}{2^{n}}

The Attempt at a Solution

I don't understand why 1 doesn't work :(

Is it a typo?

 

[JonF]

This relation starts at n=0, so your induction will need to start at 0

n=0; 1/2⁰ = 2 - 1/1⁰ = 1
n=1; 1/2⁰ + 1/2¹ = 2 - 1/1¹ = 3/2

 

[Apphysicist]

1 does work...your series is given by the formula sum(0,n) 1/2^n [the sum from 0 to n, if that notation is unusual]. So if you have from 0 to 1, you should have two terms:

1/2^0 + 1/2^1 = 3/2
Also, that works for 2-1/2^n=2-1/2= 3/2

So then you should assume that sum is equal to 2-1/2^n and then show that it also works for 2-1/2^(n+1).

 

[Gooolati]

I thought the natural numbers didn't include 0 though?

 

[JonF]

Induction can start on any integer. For a lot of relations it just so happens to hold for only the naturals. If you want you can start it on 1, but it makes more sense to prove the relation for the largest set possible. For your base case of n=1 you should get 3/2, I edited my post to show this. You could also start your base case on 23 and prove it for all natural numbers 23 and larger, but that seems sort of silly – why not show the relation for as many numbers as possible?

 

[Fredrik]

I thought the natural numbers didn't include 0 though?

They didn't originally, but mathematicians in this century prefer to include 0.

 

[Gooolati]

Gotcha!
And thank you so much for the explanation! Helps a ton!

 

[JonF]

No prob. Just an FYI for you, some people count 0 as in the naturals, so don’t be shocked if you see it in the future!