Sum to infinity

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Small123

Just show him

0.99\dot{9} \equiv \displaystyle\lim_{N\to \infty}\displaystyle\sum_{k=1}^N \dfrac{9}{10^k}\equiv 1

"Just". I think with all of this, the discussion should really start with "what is a real number". How do we define them in terms of decimal expansions

Reply 21

sarubobo28

1/3=0.333...
(1/3)*3=0.3333...*3=0.999...
but (1/3)*3 =1 so 0.999...=1

But I guess whether you agree that 0.999...=1 depends on how you define infinity...

I've always thought 0.999...=1 simply because if you want to say 1=0.999...+(1/10^∞ ), (1/10^∞ ) does not exist as 0.999... continues to infinity (never ends) which means (1/10^∞ )=0, and so 0.999...=1.

I don't know if that makes sense but that was what I came up with...

So I think this proof (?) is valid:

n=0.999...
10n=9.999...
10n-n=9n=9.0
n=1

Multiplying n by 10 does not shift the "last 9" up a decimal place, simpy because the "last 9" does not exist.

Reply 22

Oh I Really Don't Care

Life is easier when you define what 'rules' real numbers must obey and properties of sets of real numbers. Talking about the 'last digit' and throwing around infinite sums willy nilly is a biy of a mathematical faux pas.

Reply 23

Bobifier

SimonM

That "proof" suggests that we have a well defined set of operations on infinite decimal expansions, which for some people (perhaps your maths teacher included) is not a trivial assumption.

I was under the impression arithmetic was defined for all real and complex numbers. Even if not, it works for all other varieties of reccuring decimal expansion...
And equally, there is the proof by considering 1/3 = 0.333333... 3*0.33333... = 0.9999999... = 3/3 = 1.

Reply 24

Planto

Yay, meaningless rhetoric.

Reply 25

Oh I Really Don't Care

Bobifier

Things aren't so nice when introduce infinty though - for instance the sum

\displaystyle \sum \frac{(-1)^n}{n}

has different values regarding the order you sum the terms - yet if this was a fininte sum this would not be the case.

You can't really say anything "works" for an infinte decimal expansions until you have prroved such an expansions is well defined and any operation you happen to do with it is well defined.

We clearly have differing interpretations of proof.

Reply 26

Oh I Really Don't Care

Dadeyemi

If the limit exists, it is a number that arithmetic works fine on, this limit clearly does not exist, but I see the point you are trying to demonstrate.

erm yeah it does exist.

Reply 27

Simplicity

I don't like these threads.

But, some numbers you can't do that trick. Certainly, square root of 2 and say pi.

Although, you can write both as a infinite sum in finite terms i.e. using notation. But, some numbers you can't do that i.e. undescribable numbers.

P.S. Anyway, 0.9999...=1

Reply 28

Dadeyemi

DeanK22

erm yeah it does exist.

thought you wrote (-1)^n :p:

Reply 29

P → Q, ¬Q ⊢ ¬P

Does there exist a real number x such that

0.9999.... < x < 1

Reply 30

DFranklin

SimonM

"Just". I think with all of this, the discussion should really start with "what is a real number". How do we define them in terms of decimal expansions.

This. Won't happen, though.

Reply 31

munn

Original post by P &#8594

Does there exist a real number x such that

0.9999.... < x < 1

Of all the methods of trying to prove this, looking at it this way makes the most sense

Reply 32

sarubobo28

Original post by P &#8594

Does there exist a real number x such that

0.9999.... < x < 1

munn

Of all the methods of trying to prove this, looking at it this way makes the most sense

One can argue that there exists no x because 0.999... is adjacent to 1 so I don't see any convincing element in this proof to show 0.999...=1. Isn't this just back to square one?