Sequences question help?

Question

For the sequence An=n/(n+1) (n belongs to the group of natural numbers), and each e>0. Find N(e) belonging to the group of natural numbers such that,

for all n greater than or equal to N(e), IAn - 1I < e.

I have never understood what this type of question is asking and where to start.

Answer 1

I have no idea what Kiwi is talking about.



Notice that the limit as n approaches infinity, of the sequence An, is 1.



|An - 1| is the difference between a sequence term and the limit.



To say | An - 1 | < e, we are saying that the difference between An and its limit is less than a particular value 'e'. We want the error between An and its limit to be below a certain maximum value e. We get to specify that e.



N(e) is a function which takes the maximum error e as its domain, and it returns as an output for a range a natural number... this natural number is your minimum n which ensures that An and all terms after it are within e of the sequence limit.



Never mind. My explanation is probably more complicated.

Answer 2

The question is highlighting the approach of showing that the sequence An converges (to 1) by saying that:



If you give me any e>0, I can find a big number N(e) for that e, such that, after N(e) terms in the sequence, the terms will all be within e from the limit, in this case 1.



That is how convergence of sequences seems to be defined.





So, the sequence is 0, 0.5, 0.66... , 0.75, 0.8, 0.833... etc

If the given e was 0.4, you could let N(0.4) = 2. Then after the 2nd term, all terms are within 0.4 of the limit, which is 1.



If the given e was 0.3, you could use N(0.3) = 3.



So, what you want to do is to find a general expression N(e) that bounds the tail of the sequence to being near enough to 1.







For any n:

|An - 1| = |n/(n+1) - 1|

= 1/(n+1)



If you want this to be less than e, you require that n+1 is greater than 1/e



So, define N(e) = the first natural number greater than 1/e.

This can be called the roof function for 1/e.



Then, we find that, if n is greater or equal to N(e):

=> n > 1/e

=> n + 1 > 1/e

=> 1/(n+1) < e

<=> |An - 1| < e since we earlier showed that |An - 1| = 1/(n+1)



Which is the implication required.

Answer 3

oh ok.



this is a modulus question using the "e" function which is basically the inverse of natural long ln.

those two lines mean they want the negative answer and positive answer, which means opposite signs, instead of - they want + also.



you can find this on the net, im sure theres a logarithm law stating e cannot be bigger than 0 or is smaller than 0 etc etc.

so this question maybe a trick question.



hope i helped, sorry couldnt say anymore, just im doing this stuff now :/