# Rigorous proof of limits of sequences (3)

1. Jan 8, 2010

### kingwinner

1. The problem statement, all variables and given/known data

But I think the definition is as follows:
Let an be a sequence of real numbers. Then an->a iff
for ALL ε>0, there exists an integer N such that n≥N => |an - a|< ε.

The definition says that it has to be true for ALL ε>0, but in the example above, they just let ε to be a rational number with a very specific form. To me, the proof looks incomplete. They only proved the statement for the case of ε being rational number with that very specific form, but how about the case when we're given an irrational ε, or other rational ε that cannot be expressed in that specific form?
Is the proof correct or not?

2. Relevant equations
N/A

3. The attempt at a solution
Shown above.

Can someone explain this? Thanks for any help! :)

Last edited: Jan 8, 2010
2. Jan 8, 2010

### snipez90

Well I'm assuming k is a real number, in which case there is nothing wrong with the proof (verify this on your own). However, there is a much simpler estimate that could have been made, namely

$$\frac{1}{n+1} \leq \frac{1}{N+1} < \varepsilon$$

provided that $N > \frac{1}{\varepsilon}-1.$

Out of curiosity, which text are you learning from?

3. Jan 8, 2010

### kingwinner

I looked at my book again and the context before the example highly suggests that k is a natural number (or 0). It says "it suffices to consider only values for ε of the form (1/2)10-k...agree with L to at least k decimals of accuracy."

If this is the case, is the proof still correct? If so, why?

(I understand your much simpler proof, but I just want to know whether "the proof provided above" is complete or not.)

The book from which this example is taken is Real Analysis by Donsig.

4. Jan 8, 2010

### snipez90

Ah OK, well the author is choosing not to consider all real epsilon, which is weird. I would say it doesn't matter much since if we let epsilon -> 0 we can still make our sequence arbitrarily close to 1. Anyways it seems simpler to consider all real values of epsilon (> 0) and replace $\frac{1}{2}10^{-k} = \varepsilon$ with $\frac{1}{2}10^{-k} < \varepsilon.$

5. Jan 9, 2010

### HallsofIvy

Staff Emeritus
Given any $\epsilon> 0$, there exist k such that $(1/2)10^{-k}< \epsilon$. Then if $|a_n- L|< (1/2)10^{-k}$, it is less than $\epsilon$ so this is sufficient.