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?