Aren't there infinitely many primes?

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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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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.
 
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.
 
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 ##.
 
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, ...
 
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.
 
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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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!