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# Solution: epsilon proof: lim 0.9999...=1

Posted Jul20-12 at 02:03 PM by John Creighto
Updated Jul20-12 at 06:19 PM by John Creighto

In another form I presented a proof (post #145)that the lim of 0.9999....=1 using epsilon delta arguments. I was asked to explain each step so I am moving it here for the latex support.

I start with what DanLanglois presents presents in post #139

Quote:
0.999... + 0.1 = 1 + 0.099...
0.999... + 0.01 = 1 + 0.0099...
0.999... + 0.001 = 1 + 0.00099..
I do not prove this but it follow from basic arithmetic (by the way numbers carry in addition). One could prove this using mathematical induction.

For instance:
Code:
 0.01
+0.99999...
---------
=0.1
+0.00
+0.90999...
----------
=1
+0.0
+0.00999...
----------
1.099999...

The next fact that we use is that for two series series {a_n} and {b_n}, if a_n=b_n it is true then:

$$\lim_{n\to\infty}\{b_n\}=\lim_{n\to\infty}\{b_n\}$$

I claim that:

$$\lim_{n\to\infty}\left(0.999... + 10^{-n}\right) = lim_{n\to\infty}\left(0.999...\right)$$

This is straight forward by showing the difference of these two series converges to zero. With this fact it is sufficient to show that:

(1) -- $$\lim_{n\to\infty}\left(1+\sum_{n}^{\infty}9*10^{-n} \right) = 1$$
(note - we could have got to this point instead by using the geometric series as DanLanglois suggestes in post #146)

Now notice that 1>0.99, 1>0.999 etc, (which is provable by induction). Using this fact we get the inequality:

(2) -- $$\left(1+\sum_{n}^{\infty}9*10^{-n} \right) < \left(1+10^{-(n-1)} \right)$$

subtracting 1 from each side:

(3) -- $$\left(\sum_{n}^{\infty}10^{-n} \right) < \left(10^{-(n-1)} \right)$$

This fact will be used later.

Equation (1) is the limit we are trying to prove. We need to show that for any possible ε there exist an N so that for all n>N it is true that.

(4) -- $$\left|\left(1+\lim_{n\to\infty}\sum_{n}^{\infty}10^{-n} \right) - 1 \right| < \epsilon$$

This equation can be simplified by bringing the 1 inside the limit. (again this follows from the linearity of the limit.)

(5) -- $$\left| \left(\lim_{n\to\infty}\sum_{n}^{\infty}10^{-n} \right) \right| < \epsilon$$

To prove this we need to show that for any possible ε there exist an N so that for all n>N equation (5) is true.

We start with the left hand side of the inequality and apply inequality (3)

(6) -- $$\left| \left( \sum_{n}^{\infty}10^{-n} \right) \right| < \left| \left(10^{-(n-1)} \right) \right|$$

now we want to make the thing on the left less then ε for any n>N. Well, what if the thing on the left was equal to epsilon for n=N

(7) -- $$\left( 10^{-(N-1)} \right)= \epsilon$$

then because $$10^{-(n-1)}$$ is monotonically decreasing it follows for n>N that:

(8) -- $$\left(10^{-(n-1)} \right)<\left(10^{-(N-1)} \right)$$

using this in (6) we get:

(9) -- $$\left| \left( \sum_{n}^{\infty}10^{-n} \right) \right| < \left| \left( 10^{-(N-1)} \right) \right|=\epsilon$$

which is what we want. QED.
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