# Is the converse of this theorem true or not?

Byeonggon Lee
Homework Statement:
Is converse of this theorem true or not?
Relevant Equations:
$$\lim_{n\to\infty} a_n$$
This theorem is from the stewart calculus book 11.1.6
If $$\lim_{n\to\infty} |a_n| = 0$$, then $$\lim_{n\to\infty} a_n = 0$$
I wonder whether converse of this theorem true or not

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Homework Helper
I would say yes.
In fact, that looks more like a definition for "limit" as it applies to complex numbers than it does as a theorem.

Homework Helper
Gold Member
2022 Award
Problem Statement: Is inverse of this theorem true or not?
Relevant Equations: $$\lim_{n\to\infty} a_n$$

This theorem is from the stewart calculus book 11.1.6
If $$\lim_{n\to\infty} |a_n| = 0$$, then $$\lim_{n\to\infty} a_n = 0$$
I wonder whether inverse of this theorem true or not

I think "converse" is the correct terminology.

Can you prove it?

What if the limit is non zero?

Byeonggon Lee
I think "converse" is the correct terminology.
Yes you're right. I edited the post.
converse of the theorem:
$$\lim_{n\to\infty} a_n = 0 \Longrightarrow \lim_{n\to\infty} |a_n| = 0$$
I tried to prove it by using contrapositive
contrapositive of the converse:
$$\lim_{n\to\infty} a_n \neq 0 \Longrightarrow \lim_{n\to\infty} |a_n| \neq 0$$
I think if $$\lim_{n\to\infty} a_n = c$$ $$(c \neq 0)$$
this will be also true by limit laws for sequences:
$$\lim_{n\to\infty} |a_n| = |c|$$ $$(c \neq 0)$$
But I don't know what to do when
$$\lim_{n\to\infty} a_n = \infty$$

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Homework Helper
Gold Member
2022 Award
Yes you're right. I edited the post.
converse of the theorem:
$$\lim_{n\to\infty} a_n = 0 \Longrightarrow \lim_{n\to\infty} |a_n| = 0$$
I tried to prove it by using contrapositive
contrapositive of the converse:
$$\lim_{n\to\infty} a_n \neq 0 \Longrightarrow \lim_{n\to\infty} |a_n| \neq 0$$
I think if $$\lim_{n\to\infty} a_n = c$$ $$(c \neq 0)$$
this will be also true by limit laws for sequences:
$$\lim_{n\to\infty} |a_n| = c$$ $$(c \neq 0)$$
But I don't know what to do when
$$\lim_{n\to\infty} a_n = \infty$$
You need to be careful. First, the negation of the statement that ##a_n## converges to ##0## is that it does not converge to ##0##. This includes cases where the limit does not exist.

What you have for the contrapositive of the converse is not right, therefore.

But, in any case, you should be able to prove directly that the convergence of ##a_n## to ##0## and the convergence of ##|a_n|## to ##0## are equivalent. Just use the definition of the limit.

You also need to be more careful if the limit is nonzero and should look at the problem carefully.

Note that in this case the convergence of the sequence and the convergence of the absolute value are not equivalent.

The same goes for sequences that diverge to ##\pm \infty##.

sysprog and Byeonggon Lee
Gold Member
If ##|a_n - 0| < \epsilon##, then also ##||a_n |-0|<\epsilon##, and the other way around. It's just about taking absolute value one or two times.

sysprog and Byeonggon Lee
Trivially true using epsilon- definition. Use that ##||x||=|x|##.

sysprog and Byeonggon Lee
Byeonggon Lee
Thanks. Definitely better to use the definition than contrapositive.

sysprog
Homework Helper
the fact that the absolute value is continuous implies the converse is true. More interesting is the fact that the theorem itself is not always true. I.e. the absolute values lie in the real numbers which is a complete field, but the original points may lie in a non complete space, hence the absolute values may converge in the reals, without the original points themselves converging in the original space. E.g. let the original space be the rationals. Then one can have the absolute values of a sequence of rationals converging in the reals to an irrational number, whence the original points do not converge in the rationals. In general absolute convergence of a sequence of points of a normed vector space implies convergence of the sequence itself if and only if cauchy convergence implies convergence, i.e. completeness holds. I hope this is correct; it has been over 50 years since I learned this from Lynn Loomis, but it is pretty firm in memory.

a reference is Advanced Calculus by Loomis and Sternberg, Theorem 7.11 page 221, and exercise 7.19, p. 223.

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Byeonggon Lee and sysprog