Definition of a limit of a sequence

[lizarton]

Homework Statement

Use the definition of a limit to prove that lim [(1+a_n)^-1] = 1/2 if lim a_n = 1.

Homework Equations

($\forall$$\epsilon$>0)($\exists$N$\in$N)(n$\geq$N $\Rightarrow$|a_n-L|<$\epsilon$)

The Attempt at a Solution

Let $\epsilon$ be arbitrary. Since lim a_n exists, $\exists$N$\in$N such than |a_n-1|<$\epsilon$'.

My professor helped me a bit, but once we started comparing two different epsilons, I couldn't follow him anymore. He said to choose $\epsilon$'< 1/2 since 1/2 < a_n, but I don't understand why we can say that the sequence is greater than or equal to 1/2 since we only know the value of its limit.

Any help would be appreciated, I've always had a hard time with the rigorous definitions.

 

[SammyS]

Welcome to PF.

Let me try to read his mind.

Since lim a_n exists, ∃N∈ℕ such than |a_n-1| < ϵ'.

Letting ϵ' = 1/2, since it is arbitrary, gives: |a_n-1| < 1/2  → -1/2 < a_n-1 < 1/2

Therefore, 1/2 < a_n < 3/2.  It's the 1/2 < a_n you're interested in.


If a_n > 1/2, then 2(a_n + 1) > 3  → $\displaystyle \frac{1}{2(a_n+1)}<\frac{1}{3}$
​

That should be handy in working with |(1+a_n)^-1 - 1/2|, when you write that as one fraction with a common denominator.

 

[lizarton]

So then, we have showed that 1/(1+a_n) < epsilon'/3 < epsilon', but is this the end of the proof? I guess my question is, when does the other epsilon come in? or is epsilon' just our other epsilon in disguise? Thank you so much! I could use a mind-reader for this professor :)

 

[lizarton]

Also, if you wouldn't mind--(this homework is due in an hour, so this is more for my understanding)--in general, how do we know how to choose epsilon so that it will give us the result we are looking for? Is there a reason choosing epsilon'=1/2 was particularly useful for this problem?

 

[SammyS]

So then, we have showed that 1/(1+a_n) < epsilon'/3 < epsilon', but is this the end of the proof? I guess my question is, when does the other epsilon come in? or is epsilon' just our other epsilon in disguise? Thank you so much! I could use a mind-reader for this professor :)

No, that's not the complete proof.

Let ε' = min(ε, 1/2)

Then you have N∊ℕ such that for n > N, |a_n-1| < ε' ≤ 1/2

This leads to 1/(2|a_n+1|) < 1/3.

Also, for n > N, |a_n-1| < ε' ≤ ε

Put them together, & you get (the desired quantity) < ε/3 < ε​