# How do you prove that if $\lim\limits_{n \to \infty}a_n=1$, then $\lim\limits_{n \to \infty}\frac{1}{1+a_n}=\frac{1}{2}$?

More precisely:

Prove using only the $$\epsilon$$-$$N$$ definition of convergence that if $$\lim\limits_{n \to \infty}a_n=1$$ and $$a_n>-1$$ for all $$n\in \mathbb{N}$$, then $$\lim\limits_{n \to \infty}\frac{1}{1+a_n}=\frac{1}{2}$$ .

Here's what I have so far:

1. Let $$\{a_n\}$$ be a sequence and suppose $$\lim\limits_{n \to \infty}a_n=1$$ and $$a_n>-1$$ for all $$n\in \mathbb{N}$$.
2. Then for all $$\epsilon>0$$, there exists $$N\in \mathbb{N}$$ such that for all $$n\ge N$$, $$|a_n-1|<\epsilon$$ by the $$\epsilon$$-$$N$$ definition of convergence.
3. Then $$-\epsilon
4. Then $$-\epsilon<1+a_n-2<\epsilon$$
5. Then $$\frac{1}{-\epsilon}<\frac{1}{1+a_n}-\frac{1}{2}<\frac{1}{\epsilon}$$
6. Then $$|\frac{1}{1+a_n}-\frac{1}{2}|<\frac{1}{\epsilon}$$
7. Let $$\epsilon'=\frac{1}{\epsilon}$$
8. Then for all $$\epsilon'>0$$, there exists $$N\in \mathbb{N}$$ such that for all $$n\ge N$$, $$|\frac{1}{1+a_n}-\frac{1}{2}|<\epsilon'$$
9. Therefore, $$\lim\limits_{n \to \infty}\frac{1}{1+a_n}=\frac{1}{2}$$ by the $$\epsilon$$-$$N$$ definition of convergence.

Is this a valid proof? In particular, I am not sure about step 5. Intuition tells me that it is correct; but I am not 100% sure about the algebra.

• How do you get from step 4. to step 5.? This should be wrong. – Cornman Jul 14 at 1:44
• It goes wrong after step 5. First, you cannot obtain 5 from 4 (4 is right). Second, if $\epsilon$ goes to $0$, then $\epsilon'$ goes to $\infty$ so it is useless in the proof. Hint: you are right to compare $\frac{1}{1+a_n}-\frac{1}{2}$. Write the expression and then think about how to make use of 4. – trisct Jul 14 at 1:45

You should start with $$\frac{1}{1+a_n}$$ rather than $$a_n$$. Here is a standrad answer.
For any $$\epsilon>0$$, we want to find an $$N$$ such that for all $$n>N$$, $$\left|\frac{1}{1+a_n}-\frac12\right|=\frac{|2-(1+a_n)|}{2(1+a_n)}=\frac{|1-a_n|}{2(1+a_n)}<\epsilon.$$ Since $$\lim a_n=1$$, there exists an $$N_1$$ such that $$a_n>0$$ for any $$n>N_1$$. Choose $$N_2$$ such that $$|1-a_n|<2\epsilon$$ for any $$n>N_2$$. Let $$N=\max{(N_1,N_2)}$$, then for any $$n>N$$, we have $$\frac{|1-a_n|}{2(1+a_n)}<\frac{2\epsilon}{2}=\epsilon.$$ From the definition we finally prove the desired result.
The five step is wrong. Better before you notice that $$\left| \frac{1}{1+a_n} - \frac{1}{2} \right| = \frac{1}{2} \left| \frac{a_n-1}{a_n+1} \right|$$ now, the factor $$| a_n-1 |$$ is already bounded for some $$\varepsilon$$. Just missing find an upper bound for the term $$\frac{1}{| a_n+1 |}$$ (as advice, use all your hypotheses)