Use infinite series to show 0.9999999999 (repeating) =1?

Question

please help me.

Answer 1

Well, first we need to find the decimal representation of this number so it can be broken up into several terms. Well, you could add up the first nine, which is 0.9, and then add 0.09, then add 0.009, and so on... In other words,

0.9999999... = 0.9 + 0.09 + 0.009 + 0.0009...

You'll notice that each subsequent term is equal to 1/10 the previous term, so you have an infinite geometric series in which you can apply the convergence theorem:

∞
Σ [ a r^n ] = ar + ar² + ar³ + ... = (a r)/(1 - r)
n=1

In this case, the first term would be 9*(1/10), the second term is 9*(1/10)², and so on... Therefore, in this case, our 'a' is 9, and 'r' is 1/10. So, we'll just plug in our numbers:

∞
Σ [ 9 (1/10)^n ] = 9(1/10) + 9(1/10)² + 9(1/10)³ + ...
n=1

∞
Σ [ 9 (1/10)^n ] = (9*(1/10)) / (1 - 1/10) = 0.9 / 0.9 = 1.
n=1

Q.E.D.

Answer 2

This is the series 9/10 + 9/100 + 9/1000 + ... This is a geometric series with a (first term) = 9/10 and r (ratio) = 1/10. the sum of such a series is known to be a/(1 - r), whenever r is less than 1 in absolute value. Thus the sum is
(9/10)/(1 - 1/10) = (9/10)/(9/10) = 1.

Answer 3

The partial fraction decomposition is a million/(n(n+a million)(n+2)) = (a million/2) (a million/n) - a million/(n+a million) + (a million/2) a million/(n+2) there is a few cancellation once you start up computing partial sums despite the fact that it takes only a splash artwork to work out. enable s_k denote the kth partial sum. s_1 = (a million/2) [a million] - a million [a million/2] + (a million/2) [a million/3] s_2 = s_1 + (a million/2) [a million/2] - a million [a million/3] + (a million/2) [a million/4] s_3 = s_2 + (a million/2) [a million/3] - a million [a million/4] + (a million/2) [a million/5] in case you calculate s_3 case in point, and assemble the coefficients of the numbers in sq. brackets mutually, you will see that [a million/3] does no longer look using fact the (a million/2) [a million/3] and the (a million/2) [a million/3] in s_3 are canceled out by potential of the -[a million/3] in s_2. it extremely is the trend that repeats. once you calculate s_k, in addition, for many m between a million and ok, there will be a a million/2 (a million/m) term in s_{m-2}, a a million/m in s_{m-a million}, and a -a million/2 (a million/m) in s_m, and that they cancel completely. the only issues that don't cancel out completely are the few words on the beginning up (concerning [a million] and [a million/2] and the few on the tip (concerning [a million/(ok+a million)] and [a million/(ok+2)]. in case you shop careful song of this you detect s_k = (a million/2) a million - a million (a million/2) + (a million/2) (a million/2) - (a million/2) a million/(ok+a million) + (a million/2) a million/(ok+2) greater only s_k = a million/4 - (a million/2) [a million/(ok+a million) - a million/(ok+2)] = a million/4 - (a million/2) (a million/((ok+a million)(ok+2))), a formulation you would be sure by potential of induction. observe that it does certainly circulate to a million/4 as ok is going to infinity.

Answer 4

Can't be done. It is clear .9999999.......< 1.0000000000 no matter how close you take it.

What you are referring to is a supposed proof that .9999999.... = 1.0000000...... In fact there are several approaches to do this, but all of them are mathematically flawed. [See source.]

Check out this "proof" that shows any number is equal to any number...something like proving .99999...... = 1.0........

Let t = .999999999.... and a + b = 1.0........

a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b [See source.]

In other words a + b can equal any number t and yield a = b or a + b = 2b. This fallacious proof results because we've multiplied both side of the equation by zero (a - b), which we see from a = b, which means a - b = 0. Thus a - b = 1/infinity. This is similar to the .999999 ... = 1.000.... proof where we are dealing with infinity, in this case an infinite number of decimal points.

Even math teachers have been known to prove the point nines = 1.000 But as I said, mathematicians, even advanced mathematicians, do not handle inifinity very well...it is a fundamental flaw in our basic tenets of mathematics.

This will not be a popular answer because the mathematicians swear by the convergence on a limit as the number of decimal points goes to infinity. And I agree, the limit does approach 1.000, but, and this is a big BUT, it never really reaches 1.000..... no matter how many decimal points we take this.

Answer 5

Let's have a look at the sum from j=1 to n of 9*10^-n

1 - sum from j=0 to n of 9*10^-j = 0.1^n

For every d>0 we can find n large enough such that

1 - sum from j=0 to n of 9*10^-j = 0.1^n < d

0.1^n < d <=> 10^n * 0.1^n < 10^n * d <=>
<=> 1 < 10^n * d <=> 10^n > 1/d <=>
<=> log(10^n) > log(1/d) <=> n > -log d

QED

Answer 6

You don't need an infinite series to prove it. The proof follows from the definition of real numbers as a continuum:

"If there are two numbers x and z such that x < z, then there exists a third number y, such that x < y < z"

which says in plain English "If two numbers are different, there is always another number in-between them."

What number is between 0.99999999~ and 1? There is no such number, therefore the two numbers are the same.

Answer 7

i don't really know what infinite series but heres how i would do it

.9999999999..... = .9 + .09 + .009 + .0009

each term is multiply by 1/10

use the geometric sum fomula:

sum = a1 / (1 - r)
sum = .9 / (1 - 1/10)
sum = .9 / (.9
sum = 1

Answer 8

=0.9( 1+1/10+1/100++++) = 0.9(1/(1-1/10)=0.9(10/9)=1

Answer 9

let x = .9999(repeating)
then 10x = 9.9999 (repeating)

10x - x = 9.999(repeating) - .9999(repeating) = 9.0
10x - x = 9x
so 9x = 9, therefore divide both sides by 9, x = 1.

but we started with x= .9999(repeating)

therefore .9999(repeating) equals 1 Q.E.D.

Answer 10

to add to this discussion:

1/3 = .33333.....
+2/3 = .66666.....
-----------------------
3/3=1=.9999999...