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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?

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    $\begingroup$ 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. $\endgroup$ – rtybase Oct 6 '18 at 19:32
  • $\begingroup$ @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? $\endgroup$ – dc3rd Oct 6 '18 at 19:33
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    $\begingroup$ 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 "$\forall \epsilon > 0 \ \exists \ N_2 \in$" $\endgroup$ – rtybase Oct 6 '18 at 19:36
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    $\begingroup$ 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 $1/n < \epsilon/2$. $\endgroup$ – GEdgar Oct 6 '18 at 19:36
  • $\begingroup$ You should specify somewhere that $(a_n)_n$ converges, it is not explicitly written as a statement in the initial wording. $\endgroup$ – zwim Oct 6 '18 at 20:21
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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!

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