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

limitsproof-verificationreal-analysis

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 $-\epsilon<a_n-1<\epsilon$
  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.

Best Answer

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.

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