Aren't there infinitely many primes?

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In summary, the sequence "a_n" tends to a limit as n goes to infinity, while the sequence "\frac{1}{primenumber}" does not have a limit.
  • #1
PcumP_Ravenclaw
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4

Homework Statement


sn= 1/n if n is a prime number; sn = 0 if n is not prime.

Homework Equations

The Attempt at a Solution


We know that there are infinitely many prime numbers as n tends to infinity so why is ## \frac{1}{(prime number)} ## not equal to zero when n is a prime number?
 
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  • #2
PcumP_Ravenclaw said:

Homework Statement


sn= 1/n if n is a prime number; sn = 0 if n is not prime.

Homework Equations

The Attempt at a Solution


We know that there are infinitely many prime numbers as n tends to infinity so why is ## \frac{1}{(prime number)} ## not equal to zero when n is a prime number?
It looks like Sn is a term in a series with a particular definition. That definition is what you have given.

Is this a fragment from a problem you have been given.

Please state the complete problem.
 
  • #3
SammyS said:
It looks like Sn is a term in a series with a particular definition. That definition is what you have given.

Is this a fragment from a problem you have been given.

Please state the complete problem.

The question to this problem is as below.

For each of the sequences defined in below, state whether or not it tends
to a limit. If a sequence has a limit, make an ##(\epsilon,N )## table, taking
## \epsilon = 0.001 ## and any other values that you like.
 
  • #4
PcumP_Ravenclaw said:

Homework Statement


sn= 1/n if n is a prime number; sn = 0 if n is not prime.

Homework Equations

The Attempt at a Solution


We know that there are infinitely many prime numbers as n tends to infinity so why is ## \frac{1}{(prime number)} ## not equal to zero when n is a prime number?

When n is an integer (prime or not), 1/n is not zero; however, it is small if n is large. In fact, 1/n is never 0, but ##0 = \lim_{n \to \infty} 1/n ##.
 
  • #5
Because there are an infinite number of primes, the limit as n goes to infinity of your sequence is 0. Now, [itex]\frac{1}{primenumber}[/itex] is a different sequence all though its limit is also 0.

Your series "[itex]a_n[/itex]" is 0, 1/2, 1/3, 0, 1/5, 0, 1/7, 0, 0, 0, 1/11, ...
Your series "[itex]\frac{1}{primenumber}[/itex]" is 1/2, 1/3 ,1/5, 1/7, 1/11, 1/13, 1/17, ...
 
  • #6
HallsofIvy said:
Because there are an infinite number of primes, the limit as n goes to infinity of your sequence is 0. Now, [itex]\frac{1}{primenumber}[/itex] is a different sequence all though its limit is also 0.

Your series "[itex]a_n[/itex]" is 0, 1/2, 1/3, 0, 1/5, 0, 1/7, 0, 0, 0, 1/11, ...
Your series "[itex]\frac{1}{primenumber}[/itex]" is 1/2, 1/3 ,1/5, 1/7, 1/11, 1/13, 1/17, ...
how do you know that for the an sequence non-prime terms for n equal 0? its not given in the question.
 
  • #7
PcumP_Ravenclaw said:
how do you know that for the an sequence non-prime terms for n equal 0? its not given in the question.
It is given in the definition of the sequence.

It seems you don't understand the word "defined".

PcumP_Ravenclaw said:
The question to this problem is as below.

For each of the sequences defined in below, state whether or not it tends
to a limit. If a sequence has a limit, make an ##(\epsilon,N )## table, taking
## \epsilon = 0.001 ## and any other values that you like.
PcumP_Ravenclaw said:

Homework Statement


sn= 1/n if n is a prime number; sn = 0 if n is not prime.
...
The definition of this sequence is:

sn = 1/n, if n is prime
sn = 0, if n is not prime​

1 is not prime, so s1 = 0
2 is prime, so s2 = 1/2
3 is prime, so s3 = 1/3
4 is not prime, so s4 = 0
5 is prime, so s5 = 1/5
6 is not prime, so s6 = 0
7 is prime, so s7 = 1/7
8 is not prime, so s8 = 0
9 is not prime, so s9 = 0
10 is not prime, so s10 = 0
11 is prime, so s11 = 1/11
...

Get it ?
 
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  • #8
SammyS said:
It is given in the definition of the sequence.

It seems you don't understand the word "defined".
The definition of this sequence is:

sn = 1/n, if n is prime
sn = 0, if n is not prime​

1 is not prime, so s1 = 0
2 is prime, so s2 = 1/2
3 is prime, so s3 = 1/3
4 is not prime, so s4 = 0
5 is prime, so s5 = 1/5
6 is not prime, so s6 = 0
7 is prime, so s7 = 1/7
8 is not prime, so s8 = 0
9 is not prime, so s9 = 0
10 is not prime, so s10 = 0
11 is prime, so s11 = 1/11
...

Get it ?
yes right. Its given in the question. danke! Haste leads to waste!
 

1. What does it mean for there to be infinitely many primes?

Being infinitely many means that there is no specific number or limit to how many primes exist. In other words, there is always another prime number that can be found, no matter how high you count.

2. How do we know that there are infinitely many primes?

This is a well-known mathematical theorem called Euclid's Theorem, which states that for any finite list of prime numbers, there is always another prime number that is not on that list. This essentially proves that there are infinitely many primes.

3. Is there a pattern to the distribution of prime numbers?

There is no known pattern or formula for the distribution of prime numbers. They appear to be randomly scattered throughout the number line, making it difficult to predict or find a pattern.

4. Can we find all the prime numbers that exist?

No, it is impossible to find all the prime numbers that exist because there are infinitely many of them. We can continue to find more and more primes, but we will never reach a point where we have found them all.

5. Why are prime numbers important in mathematics?

Prime numbers are important because they are the building blocks of all other numbers. Every positive integer can be expressed as a unique combination of prime numbers. They also have many applications in cryptography, number theory, and other areas of math and science.

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