Aren't there infinitely many primes?

[PcumP_Ravenclaw]

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?

 

[SammyS]

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 S_n 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.

 

[PcumP_Ravenclaw]

It looks like S_n 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.

 

[Ray Vickson]

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 $0 = \lim_{n \to \infty} 1/n$.

 

[HallsofIvy]

Because there are an infinite number of primes, the limit as n goes to infinity of your sequence is 0. Now, \frac{1}{primenumber} is a different sequence all though its limit is also 0.

Your series "a_n" is 0, 1/2, 1/3, 0, 1/5, 0, 1/7, 0, 0, 0, 1/11, ...
Your series "\frac{1}{primenumber}" is 1/2, 1/3 ,1/5, 1/7, 1/11, 1/13, 1/17, ...

 

[PcumP_Ravenclaw]

Because there are an infinite number of primes, the limit as n goes to infinity of your sequence is 0. Now, \frac{1}{primenumber} is a different sequence all though its limit is also 0.

Your series "a_n" is 0, 1/2, 1/3, 0, 1/5, 0, 1/7, 0, 0, 0, 1/11, ...
Your series "\frac{1}{primenumber}" 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.

 

[SammyS]

how do you know that for the a_n 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".

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.

Homework Statement

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

The definition of this sequence is:

s_n = 1/n, if n is prime
s_n = 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 ?

 

[PcumP_Ravenclaw]

It is given in the definition of the sequence.

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

s_n = 1/n, if n is prime
s_n = 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!