Is My Solution for Limit of a Sequence Correct?

[cloud360]

! Sequences and series "limit" question, is my solution correct?

Homework Statement

[PLAIN]http://img233.imageshack.us/img233/7195/sands2010q1.gif

Homework Equations

The Attempt at a Solution

Solution posted in image above, want to know if its correct

 

[cloud360]

I am not sure how to solve when it says to find N(epsilon) such that a_n-1|<epsilon.

I am used to finding N(epsilon) for |a_n|<epsilon.

Here are some example questions and answers, my lecturer provided to |a_n|<epsilon, type questions

[PLAIN]http://img199.imageshack.us/img199/1206/sands2010q1example.gif

 

[SammyS]

Notice that n/(n+1) < 1 for all n>0 , so that |a_n - 1| = 1 - a_n .

Then notice that n/(n+1) = ((n+1)-1)/(n+1) = 1 - 1/(n+1) = a_n .

Now proceed with |a_n - 1| < ε to find N(ε) .

 

[cloud360]

Notice that n/(n+1) < 1 for all n>0 , so that |a_n - 1| = 1 - a_n .

Then notice that n/(n+1) = ((n+1)-1)/(n+1) = 1 - 1/(n+1) = a_n .

Now proceed with |a_n - 1| < ε to find N(ε) .

thanks for your reply :)

so is my solution wrong? if so where did i go wrong

 

[SammyS]

|a_n - 1| < ε is equivalent to -ε < a_n - 1 < ε

You did the inequality on the right: a_n - 1 < ε , which is true for all n since a_n < 1 for all n > 0.

The important thing is to find N(ε) such that 1- n/(n+1) < ε for n > N(ε).

The algebra may go smoother if you use a_n=\frac{n}{n+1}=\frac{(n+1)-1}{n+1}=1-\frac{1}{n+1}

 

[cloud360]

|a_n - 1| < ε is equivalent to -ε < a_n - 1 < ε

You did the inequality on the right: a_n - 1 < ε , which is true for all n since a_n < 1 for all n > 0.

The important thing is to find N(ε) such that 1- n/(n+1) < ε for n > N(ε).

The algebra may go smoother if you use a_n=\frac{n}{n+1}=\frac{(n+1)-1}{n+1}=1-\frac{1}{n+1}

|an-1|<epsilon means the limit of an=1

if i showed the limit of an=1, will that be enough to prove the first question?

thanks again for your help

 

[VietDao29]

|an-1|<epsilon means the limit of an=1

if i showed the limit of an=1, will that be enough to prove the first question?

thanks again for your help

No, that isn't enough, since the problem asks you to find N (\epsilon), such that:

n &gt; N ( \epsilon ) \Rightarrow |a_n - 1| &lt; \epsilon​

Showing that \lim a_n = 1, doesn't have anything to do with pointing out the value N (\epsilon) the problem requires. They are 2 (quite) distinct processes.

In this question, you don't need to find the limit of the sequence. Your first 4 lines are not needed.

The 5-th line is wrong because you need to solve for n such that: \left| a_n - 1 \right| &lt; \epsilon, not that a_n - 1 &lt; \epsilon

Note that, as others have pointed out, a_n is less than 1, forall n. So, can you break the absolute sign for |a_n - 1|?