Question
Prove that
1
0
1
1 , 1
1
, 1
p
if p
dx p
x
diverges if p
1
0
1
1 , 1
1
, 1
p
if p
dx p
x
diverges if p
Answers
GPT 3.5
To prove that the sequence 1, 0, 1, 1, 1, ..., p diverges if p > dx(p)/x, we can use the definition of divergence.
Definition of divergence: A sequence diverges if it does not have a finite limit.
To show that the sequence 1, 0, 1, 1, 1, ..., p diverges, we will assume the opposite, that it converges, and then derive a contradiction.
Assume the sequence 1, 0, 1, 1, 1, ..., p converges to some limit L. Then, for any positive ε, there exists some positive N such that for all n > N, |an - L| < ε, where an represents the nth term of the sequence.
Let's choose ε = dx(p)/x - p > 0. Since p > dx(p)/x, ε is positive.
By the definition of convergence, there exists some positive N such that for all n > N, |an - L| < ε.
Consider the term an = p. Since p > dx(p)/x and n > N, we have |an - L| < ε, which simplifies to |p - L| < ε.
Since ε = dx(p)/x - p, we can rewrite the inequality as |p - L| < dx(p)/x - p.
Now, let's consider the term an = dx(p)/x. Since n > N, we have |an - L| < ε.
Substituting dx(p)/x for an, we have |dx(p)/x - L| < ε.
Since ε = dx(p)/x - p, we can rewrite the inequality as |dx(p)/x - L| < dx(p)/x - p.
Combining the two inequalities, we have |p - L| < dx(p)/x - p and |dx(p)/x - L| < dx(p)/x - p.
Adding the two inequalities together, we get |p - L| + |dx(p)/x - L| < 2(dx(p)/x - p).
The left side of the inequality represents the sum of two positive quantities, which is greater than or equal to the absolute value of their difference. Therefore, we can rewrite the inequality as |p - dx(p)/x| < 2(dx(p)/x - p).
Since dx(p)/x > p, 2(dx(p)/x - p) is positive.
Now, consider the term an = 0. Since n > N, we have |an - L| < ε.
Substituting 0 for an, we have |0 - L| < ε, which simplifies to |L| < ε.
Since ε = dx(p)/x - p, we can rewrite the inequality as |L| < dx(p)/x - p.
But dx(p)/x - p is positive, so |L| < dx(p)/x - p implies that L < dx(p)/x - p.
This means that L is less than dx(p)/x - p, which contradicts the assumption that L is the limit of the sequence.
Therefore, our assumption that the sequence 1, 0, 1, 1, 1, ..., p converges must be false, and the sequence diverges if p > dx(p)/x.
Definition of divergence: A sequence diverges if it does not have a finite limit.
To show that the sequence 1, 0, 1, 1, 1, ..., p diverges, we will assume the opposite, that it converges, and then derive a contradiction.
Assume the sequence 1, 0, 1, 1, 1, ..., p converges to some limit L. Then, for any positive ε, there exists some positive N such that for all n > N, |an - L| < ε, where an represents the nth term of the sequence.
Let's choose ε = dx(p)/x - p > 0. Since p > dx(p)/x, ε is positive.
By the definition of convergence, there exists some positive N such that for all n > N, |an - L| < ε.
Consider the term an = p. Since p > dx(p)/x and n > N, we have |an - L| < ε, which simplifies to |p - L| < ε.
Since ε = dx(p)/x - p, we can rewrite the inequality as |p - L| < dx(p)/x - p.
Now, let's consider the term an = dx(p)/x. Since n > N, we have |an - L| < ε.
Substituting dx(p)/x for an, we have |dx(p)/x - L| < ε.
Since ε = dx(p)/x - p, we can rewrite the inequality as |dx(p)/x - L| < dx(p)/x - p.
Combining the two inequalities, we have |p - L| < dx(p)/x - p and |dx(p)/x - L| < dx(p)/x - p.
Adding the two inequalities together, we get |p - L| + |dx(p)/x - L| < 2(dx(p)/x - p).
The left side of the inequality represents the sum of two positive quantities, which is greater than or equal to the absolute value of their difference. Therefore, we can rewrite the inequality as |p - dx(p)/x| < 2(dx(p)/x - p).
Since dx(p)/x > p, 2(dx(p)/x - p) is positive.
Now, consider the term an = 0. Since n > N, we have |an - L| < ε.
Substituting 0 for an, we have |0 - L| < ε, which simplifies to |L| < ε.
Since ε = dx(p)/x - p, we can rewrite the inequality as |L| < dx(p)/x - p.
But dx(p)/x - p is positive, so |L| < dx(p)/x - p implies that L < dx(p)/x - p.
This means that L is less than dx(p)/x - p, which contradicts the assumption that L is the limit of the sequence.
Therefore, our assumption that the sequence 1, 0, 1, 1, 1, ..., p converges must be false, and the sequence diverges if p > dx(p)/x.