Modulus operation

Question

I was studying ratio test in infinite sequence. Then I came across this $$\left| \frac{a_{n+1}}{a_n} - l \right| < \epsilon$$ i.e. $$\left| \frac{a_{n+1}}{a_n} \right| < |l|+\epsilon$$

I don't understand how $| a-b | < c$ becomes $|a| < |b| + c$. Is that a formula?

score 2 · Answer 1 · answered Sep 25 '15 at 03:47

2

We have: $|a-b| \geq |a|-|b| \Rightarrow c > |a|-|b| \Rightarrow c+|b| > |a|$

answered Sep 25 '15 at 03:47

DeepSea

77,651

In Wikipedia, I saw $ | a-b | \geq | (|a| -|b|) |. $ – Vibhav Sep 25 '15 at 06:36
Isn't $|a-b|\geq |a|-|b|$? – Landon Carter Sep 25 '15 at 08:12
Please clarify that was it wrong or right, what I saw in Wikipedia? Or is it same? – Vibhav Sep 25 '15 at 10:13

score 0 · Answer 2 · answered Sep 25 '15 at 08:09

0

with $b_n = a_{n+1}/a_n$, you have the starting point $$|b_n - l|\leq ε $$ Triangle inequality gives \begin{align}|b_n| &\leq |b_n - l| + |l|\\ \implies |b_n| &\leq |l| + ε \end{align}

answered Sep 25 '15 at 08:09

Calvin Khor

34,903

Modulus operation

2 Answers2