# Thread: Prove 2 statements on limits of sequences are true

1. ## Prove 2 statements on limits of sequences are true

Hi, I have been told that the following two statements are true, but I don't have a clue how to actually prove it. I have been told that I can either prove it with words or with an example... :S

1.- If the sequence ( $a_n$) has a limit and the sequence ( $b_n$) is such that $a_j=b_j$ whenever $j>10^{100}$ then the sequence ( $b_j$) has a limit.
2.- Every real number r is the limit of a sequence all of whose terms are rational numbers.

2. 1. The limit of $(a_n) = L$ if and only if for every real number ε > 0, there exists a natural number N such that for every n > N we have $| a_n - L | < \varepsilon$
Well, the same goes for $(b_n)$! The "N" that we choose is just 10^100. So for n > 10^100, $(b_n) = (a_n)$, and therefore

$|b_n - L| = |a_n - L|$, which we already know is less than epsilon.

3. Originally Posted by juanma101285
1.- If the sequence ( $a_n$) has a limit and the sequence ( $b_n$) is such that $a_j=b_j$ whenever $j>10^{100}$ then the sequence ( $b_j$) has a limit.
2.- Every real number r is the limit of a sequence all of whose terms are rational numbers.
For #1. To say that $(a_n)\to L$ means that almost all (all but a finite collection) of the $a_n's$ are 'close' to $L$. Does that mean almost all of the $b_n's$ must be close to $L~?$. HOW & WHY?

For #2. Between any two numbers there is a rational number.
Suppose $\alpha$ is a real number.
For each positive integer $n$ there is a rational number $r_n$ between $\alpha~\&~\alpha+\frac{1}{n}$.
Is it true that $(r_n)\to\alpha~?$

4. Another way to look at (2): every real number can be written in decimal form: $x= A.a_1a_2a_3a_4...$

Show x is the limit of the sequence A, $A.a_1$, $A.a_1a_2$, $A.a_1a_2a_3$, etc. Do you see that those are all rational numbers?

For example, $\pi= 3.1415926...$ so it is the limit of the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, ...