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Prove: If the sequence $<a_{n}>$ converges to $b\in \mathbb{R}$, then the sequence $<|a_{n} - b|>$ converges to $0$.

Since $<a_{n}>$ converges to $b\in \mathbb{R}$, denoted by

$\lim_{n\rightarrow \infty}<a_{n}> = b$

then for every $\epsilon > 0$ there exists a positive integer $n_{0}$ such that $n > n_{0}\Rightarrow |a_{n} - b| < \epsilon \Rightarrow b - \epsilon < a_{n} < b + \epsilon$.

So:

$\lim_{n\rightarrow \infty}<|a_{n} - b|> \ \geq \ \lim_{n\rightarrow \infty}<||a_{n}| - |b|| = b - b = 0$

I am not sure if this is right, any suggestion would be greatly appreciated

Wolfy
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1 Answers1

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It is okay, but you don't need the result $a_n \to a, b_n \to b \implies a_n-b_n \to a-b$ here. Just notice that the definitions of $a_n \to b$ and $|a_n - b| \to 0$ are pratically the same. Given $\epsilon > 0$, exists $n_0 \in \Bbb N$ such that $n > n_0$ implies $|a_n - b| < \epsilon$. So $$||a_n-b| - 0| = ||a_n-b|| = |a_n-b|<\epsilon.$$

Ivo Terek
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  • ok, even if I don't need the result is the proof ok as is? – Wolfy Feb 20 '15 at 16:59
  • Looking closely, no, all that $\langle$ and $\rangle$ got in my way. It is not true that $|a_n-b| = |a_n| - |b|$, so your last expression is not valid. – Ivo Terek Feb 20 '15 at 17:01
  • But $|a_{n} - b| \geq ||a_{n}| - |b||$ therefore I and use that to say as the limit goes to infinity and use the fact the $a_{n}$ converges to $b$ and this $b - b = 0$ – Wolfy Feb 20 '15 at 17:06
  • And this states that the limit is greater than or equal to zero. Which in fact is trivial, since $|\cdot|$ only assume non-negative values. – Ivo Terek Feb 20 '15 at 17:08
  • Ok, so how do I correct my proof? – Wolfy Feb 20 '15 at 17:09
  • The easiest way is using the definition of $\lim_{n \to +\infty}|a_n-b|$, as have I done. Using direct manipulations with $\lim_{n \to +\infty} a_n=b, \lim_{n \to \infty}b=b,$ etc won't work. – Ivo Terek Feb 20 '15 at 17:11
  • are you saying your last statement in your answer above completes the proof? – Wolfy Feb 20 '15 at 17:12
  • Actually, my answer from "Given $\epsilon > 0$..." onwards is the entire proof.. – Ivo Terek Feb 20 '15 at 17:15
  • your a boss, thanks – Wolfy Feb 20 '15 at 17:18