The discussion revolves around a sequence defined such that \( s_n = \frac{1}{n} \) if \( n \) is a prime number, and \( s_n = 0 \) if \( n \) is not prime. Participants are exploring the implications of this definition in the context of limits and the behavior of the sequence as \( n \) approaches infinity.
The discussion is active, with participants providing insights into the sequence's behavior and limits. There is a recognition that the sequence has a specific definition, and some participants are seeking further clarification on the implications of this definition. Multiple interpretations of the sequence's behavior are being explored, particularly regarding the limits of the defined sequences.
There is an emphasis on the definition of the sequence and its implications for understanding limits. Participants are also noting that the problem may be part of a larger assignment, which requires stating whether the sequences tend to a limit and constructing \( (\epsilon, N) \) tables.
It looks like Sn is a term in a series with a particular definition. That definition is what you have given.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?
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.
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?
how do you know that for the an sequence non-prime terms for n equal 0? its not given in the question.HallsofIvy said: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, ...
It is given in the definition of the sequence.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.
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.
The definition of this sequence is:PcumP_Ravenclaw said:Homework Statement
sn= 1/n if n is a prime number; sn = 0 if n is not prime.
...
yes right. Its given in the question. danke! Haste leads to waste!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 ?