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

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?

• As a sketch - yes, it's a correct strategy. Probably you will need more formality and details if you plan to provide this as an answer to an exercise. Oct 6, 2018 at 19:32
• @rtybase It is not an exercise to hand in to anything, but for good practices what would have to be added to make it formally acceptable? Oct 6, 2018 at 19:33
• With the latest update, all that is left to do is to merge "This is true due to the Archimedian property of real numbers" with "$\forall \epsilon > 0 \ \exists \ N_2 \in$" Oct 6, 2018 at 19:36
• For example, you mention $\epsilon$ suddenly. Probably you should start with Let $\epsilon > 0$ be given. Then by the Archimedian property there is $N$ so that for all $n \ge N$ we have $1/n < \epsilon/2$. Oct 6, 2018 at 19:36
• You should specify somewhere that $(a_n)_n$ converges, it is not explicitly written as a statement in the initial wording.
– zwim
Oct 6, 2018 at 20:21

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!