Question

Can someone get me started on a proof to determine whether 1.999... = 2? I have a geometric series set up but I'm not sure it's setup correctly

Answer 1

Should be something like
\[1 + (1/9) + (1/9^2) + (1/9^3) + ....\]
That's where I'm having trouble, not sure how I can factor to get the a and r?

Answer 2

1.12

Answer 3

It doesn't equal 2 though?

Answer 4

@campbell_st What if I set \[1+9\sum_{n=1}^{\infty}\frac{ 1 }{ 10^n }\]

Answer 5

With those numbers it is not 2

Answer 6

Sorry that was rude. Would you please check your answer? because (1/9)/(1 - 1/9) = 1/8. yes it is supposed to be 2.

Answer 7

but here is a little quick maths to show its 2

x = 1.999....

10x = 19.999...


then

10x = 19.999...
- x = 1.999...
---------------------

9x = 18

x = 2..

so don't worry about geometric sequences or series... just use some basic algebra skills.

Answer 8

An important question. Is it 2 or isn't it? One way to answer this question is another question, If it isn't 2, how far from 2 is it? Since it is so that any value you name will eventually be greater than how close we can come by adding more '9's, we must conclude that it IS 2, and not just APPROACHING 2.

This is APPROACHING 2

1.9
1.99
1.999
1.9999
1.99999

See how we stop at finite places.

This IS 2

1.99999.....

Answer 9

I'll just use campbell's idea. Thanks everyone, I appreciate your help.

Answer 10

it just uses the idea that a recurring decimal can be written as a rational number a/b

Answer 11

No I meant your idea of just using the geometric series for 0.999 = 1, idk why the same for 1.999.. = 2 isn't working

Answer 12

so start with the 1st term as 9/10 then the common ratio is 1/10

so you are looking at

\[1 + S_{\infty} = 1 + \frac{\frac{9}{10}}{1 - \frac{1}{10}}\]

and you'll be able to prove it

Answer 13

so the series is

1 + 9/10 + 9/100 + 9/1000 +....

Answer 14

the problem was caused by the structure of your series..