Question

Prove that if the sequence An converges to L, then the sequence |An| converges to |L|.

Answer 1

"\(A_n\) converges to \(L\)" means that, for any positive \(\epsilon\), there is some \(N\) such that \(n>N\) implies \(|A_n-L|<\epsilon\).

You want to use this to show that there is some \(N\) such that \(n>N\) implies \(||A_n|-|L||<\epsilon\). It looks like you might have to use the triangle inequality at some point.

Answer 2

Yea I understand all that, I'm having trouble figuring out how to manipulate it. Do I want to start with the former of the two inequalities or the latter?

Answer 3

It seems like if I start with the first one I'd be showing that it is less than something similar to the second, but that's backwards, isn't it?

Answer 4

Ah just figured it out, there was a different part of the triangle inequality or a different way to apply it that I hadn't thought of.

Answer 5

Yeah, you eventually find that
\[||A_n|-|L||\le|A_n-L|<\epsilon\]

Answer 6

Right, thanks so much. Any idea if the converse is also true?

Answer 7

No, it's not. Consider the sequence \(A_n=(-1)^n\). For this one, \(L=\pm1\) (or "it doesn't exist"), whereas \(|A_n|=1\) tends to 1.

Are you using Rudin's book, by any chance? This problem seems very familiar.

Answer 8

No, Frank Morgan's but this problem isn't out of the book, could be one that our professor found in Rudin's book maybe?

Answer 9

And that makes sense. What's kind of tripping me up is that the converse would say that |An| converges to |L|, so you can't really know what L is itself. Oh but your An has no limit so regardless it isn't true. That makes sense haha sorry. Thanks so much.

Answer 10

Yeah, it happens to be the first problem from the chapter on sequences/series: "Prove that convergence of \(\{s_n\}\) implies convergence of \(\{|s_n|\}\). Is the converse true?"

Answer 11

You're welcome!