Sequences and Series?

Question

Does anyone hear know how to prove that 0.99999999999999999999999... = 1 with sequences and series? I used to remember but I cannot seem to find the answer anywhere. This isn't for a class or anything so please give a direct answer if possible. Thank you.

Answer 1

x = 0.9999...

10x = 9.9999...

10x - x = 9.9999... - 0.9999...

9x = 9

x = 1.

Answer 2

Well, let me take a crack since neither of the existing answers actually use sequences or series (at least none did as of the time I began typing).

First, recall that an infinite geometric series is a series of the form

SUM from 0 to infinity [a*(r^n)].

where a is a constant and r is called the ratio of the series.

There is a theorem that states that the infinite sum is equal to a/(1-r).

Think of the number 0.999999... as

0.9 + 0.09 + 0.009 + 0.0009 + ...

This is

9/10 + 9/100 + 9/1000 + 9/10000 + ...

=

(9/10)*[1 + 1/(10^1) + 1/(10^2) + 1/(10^3) + ...]

Inside the [ ] we have an infinite geometric series with a = 1 and r = 1/10. (Since each term is equal to the one before it times 1/10.

So the sum of the [ ] is equal to a/(1-r) = 1/(1-(1/10)) = 1/(9/10) = 10/9.

So we have

(9/10)*[1 + 1/(10^1) + 1/(10^2) + 1/(10^3) + ...]

=

(9/10)*[10/9]

=

1

Answer 3

You can express .9999999999999999 etc. as 9/10 + 9/100 + 9/1000 etc.

We can see that this is a geometric series, with first term or 9/10 and constant ratio of 1/10.
You can then plug it into the formula: a/(1-r)

(9/10)/(1-1/10)
=(9/10)/(9/10)
=1

Answer 4

.9999999... does not equal 1, it's limit is 1.
What is meant is that the series .9+.09+.009 ...
is a convergent series which converges to 1.
To put it in non-math terms, The sum of the
numbers .9+.09+.009+.0009 ... can be made
as close to 1 as you want, but will never reach
1, by adding enough numbers in the series.

Answer 5

0.99999999.... = 0.3333333.... * 3 = 1/3 * 3 = 1
since 0.3333333.... = 1/3
however as someone said before me it is actually not equal to 1 but its limit is. One way to prove this:
let n approach to infinity: lim (0.99999999.... )^n = 1 or 0
if 0.99999999....is really equal to 1 then the answer to this limit is 1 but then it is also a decimal so it can equal to 0. It is either undefined or it is 0 or 1. I would say that practically is 1 but mathemtically 1 is 0.9999999...'s limit.

Answer 6

0.99999999999999999999999999 is not equal to 1.
because 1=1. not 0.999999999999999999.

0.999999999999999999999 is approximately equal to 1, or,
0.999999999999999999999 is approaching 1...but not equal to 1.

unless it is stated to round off the 0.9999999999999999, or to find the nearest integer of that value, the answer would be 1.

Answer 7

9/10 + 9/(10^2) + 9/(10^3) + ...... + 9/(10^n)
= 9/10 x 1/(1 - 1/10)
= 1

/ = vinculum

Answer 8

0.9999999999999999.... is the sum of infinite geometrical progression whose first term a = 0.9 and common ratio r = 0.1 whence 0.999999999999999999.......
= a/(1 -- r) = 0.9 / (1 -- 0.1) = 0.9 / 0.9 = 1

Answer 9

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