# Thread: True or False Using Epsilon

1. ## True or False Using Epsilon

Decide if the following are true or false and provide a proof or counterexample:

Suppose that x{n} --> L.
a) For all e > 0, there exists a natural number n such that |x{n+1} − x{n}| < e
b) There exists a natural number n such that for all e > 0, |x{n+1} − x{n}| < e
c) There exists e > 0 such that for all natural numbers n, |x{n+1} − x{n}| < e
d) For all natural numbers n, there exists e > 0 such that |x{n+1} − x{n}| < e

e stands for epsilon, which is a very small number. { } are subscripts.

2. It's been a while, but I'll take a stab at it. these are outlines.

Originally Posted by letitbemww
Decide if the following are true or false and provide a proof or counterexample:
Suppose that x{n} -> L.
a) For all e > 0, there exists a natural number n such that |x{n+1} − x{n}| < e
This statement should look familiar to you.

b) There exists a natural number n such that for all e > 0, |x{n+1} − x{n}| < e
Suppose I fix n. then |x{n+1} − x{n}| is also a fixed difference. Is it (generally) true that this difference is always less than e > 0? ((an exception is the constant sequence. x{n} = L for all n))

c) There exists e > 0 such that for all natural numbers n, |x{n+1} − x{n}| < e
Suppose I fix an interval around L. Are there terms outside the interval, in general?

d) For all natural numbers n, there exists e > 0 such that |x{n+1} − x{n}| < e
First consider the terms of the sequence with the highest and lowest value. you can then choose epsilon sufficiently large enough to "wrap" around all of them. If there is no upper bound and/or lower bound, what does this tell you about the sequence?

3. Originally Posted by letitbemww
Decide if the following are true or false and provide a proof or counterexample:
Suppose that x{n} --> L.
c) There exists e > 0 such that for all natural numbers n, |x{n+1} − x{n}| < e
Convergent sequences are bounded. So $\left( {\exists B > 0} \right)\left[ {\forall n,\left| {x_n } \right| \leqslant B} \right]$.
So $\left| {x_{n + 1} - x_n } \right| \leqslant \left| {x_{n + 1} } \right| + \left| { - x_n } \right| \leqslant 2B$

4. Ok so here's what I've got:

a) True: lim x{n}=L, for any e>0 there exists N1 such that |x{n}-L|<e for all n>N1. Then |x{n+1}-L|<e for all n>N1. Also, for e/2>0, there exists N2 such that |x{n}-L|<e/2 for all n>N2. Then, |x{n+1}-x{n}|=|x{n+1}-L+L-x{n}|<=|x{n+1}-L|+|x{n}-L|=e/2+e/2=e.

b)False: If n is fixed, then |x{n+1}-x{n}| is a fixed difference. If x{n}=L, then this difference is not necessarily less than e>0.

c)True:
.
So
Can I just set e=2B then? So |x{n+1}-x{n}|<e for all n.

d)True: Can't I just say e=|x{n+1}-x{n}|+1 so that |x{n+1}-x{n}|<e?