Let $\{a_n\}_{n = 1}^{\infty}$ and $\{b_n\}_{n = 1}^{\infty}$ be two sequences of real numbers s.t $|a_n -b_n| < \frac{1}{n}$

Question

Let $\{a_n\}_{n = 1}^{\infty}$ and $\{b_n\}_{n = 1}^{\infty}$ be two sequences of real numbers s.t $|a_n -b_n| < \frac{1}{n}$ Show $\{b_n\}_{n = 1}^{\infty}$ converges to L also.

Proof Attempt

Notice $$\Bigg| a_n - b_n \Bigg| < \frac{1}{n} < \frac{\epsilon}{2}$$ This is true due to the Archimedian property of real numbers.

By definition for $\{a_n\}_{n = 1}^{\infty}$:

$\forall \epsilon > 0 \ \exists \ N_1 \in \mathbb{R} \ s.t \ \forall \ n \geq N_1 \ \Bigg|a_n - L \Bigg| < \frac{\epsilon}{2}$ and as well $\forall \epsilon > 0 \ \exists \ N_2 \in \mathbb{R} \ s.t \ \forall \ n \geq N_2 \ \Bigg|a_n - b_n \Bigg| < < \frac{1}{2} < \frac{\epsilon}{2}$

$$\therefore \Bigg|b_n - L \Bigg| \leq \Bigg| b_n - a_n \Bigg| + \Bigg| a_n - L \Bigg| < \frac{1}{n} + \frac{\epsilon}{2} < \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon$$

If we let $N = max\{N_1, N_2\}$

Correct reasoning?

Answer 1

Let $\epsilon>0$ given.

$$\lim_{n\to+\infty}a_n=L \implies$$ $$(\exists N_1\in \Bbb N)\;\; : \; (\forall n\ge N_1) \; |a_n-L|<\frac{\epsilon}{2}$$

On the other hand , as you said, $\Bbb R$ is Archimedian,

$$(\exists N_2 \in \Bbb N) \; : \; N_2\frac{\epsilon}{2}>1$$

Put $$N=\max\{N_1,N_2\}$$

then

$$(\forall n\ge N)$$ $$|b_n-L|\le|a_n-L|+|b_n-a_n|<\epsilon$$ done!