Series manipulation doubt

Question

Say that I have a series whose $n$th term is given by $a(n)$ and this converges to a limit $l$ as $n$ tends to infinity. We know that there exists some N for which $|a(n)-l|<|l|/3$. How can I manipulate this to prove that $|a(n)|\geq2 |l|/3$ for all $n>N$?

Answer 1

This follows from the triangle inequality $|x|+|y|\geq |x+y|$.

Now suppose $|a(n)|<2|l|/3$ and $|l-a(n)|<|l|/3$. By the triangle inequality we get $$|l|=|(l-a(n))+a(n)|\leq |l-a(n)|+|a(n)|<|l|/3+2|l|/3<|l|,$$ a contradiction.

Note that your initial statement that such an $N$ exists assumes that $l\neq 0$. (Your second statement is trivially true if $l=0$, however.)