Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm interested to know whether the following proposition is true or not:

Assume that:

$$\lim_{n \to \infty}a_n+\lim_{n \to \infty}b_n=c\in\mathbb{R}$$

so both $\lim\limits_{n \to \infty}a_n$, and $\lim\limits_{n \to \infty}b_n$ exists and finite.

Intuitively, I'm thinking yes, but I don't know how to justify it formally.

share|cite|improve this question
The assumption doesn't even make sense without both limits existing and being finite. On the other hand, for $c$ fixed outside of the implication, the other direction is not technically correct. You would need a $\exists c$ in front of the LHS, like "$(\exists c:\text{blah})\iff \text{blah}$." It means "there exists a $c$ such that." – anon May 8 '12 at 6:26
What exactly is the question? Are you asking how to formally justify an assumption??? Are you trying to show that $a+b$ is finite iff $a$ and $b$ are finite (the answer is yes, of course)? – copper.hat May 8 '12 at 6:31
Are you sure you're not confusing this with $\lim\limits_{n\to\infty}(a_n+b_n)=c\in\mathbb R$? – Martin Sleziak May 8 '12 at 6:35
@Amihai: As Martin said. You're confusing $\lim a_n+\lim b_n$ and $\lim(a_n+b_n)$. In your example, neither $\lim a_n$ nor $\lim b_n$ exists, but $\lim (a_n+b_n)=0$. – joriki May 8 '12 at 6:38
@AmihaiZivan: If I understand you correctly, your question has nothing to do with limits. If I have two numbers $a,b$ and $a+b$ is finite (which means it must be defined in the first place), then both $a$ and $b$ must be finite. – copper.hat May 8 '12 at 6:46
up vote 4 down vote accepted

The only way I can make sense of the question after the exchanges in the comments is that you're asking whether there exist two quantities such that their sum is finite but the quantities themselves are not finite. The answer is no.

There are basically two ways of regarding limits. In one, a limit is said to exist only if it exists in the strict sense that there is a finite number to which the sequence converges. Under this paradigm, already saying that the sum of the limits is finite implies that the limits exist and are thus by definition finite.

In another way of speaking, a limit can be said to exist and be $\infty$ or $-\infty$ if the sequence diverges to $\infty$ or $-\infty$, respectively. Under this paradigm, limits are no longer real numbers and cannot necessarily be added. While addition can be extended by defining $\infty+x=\infty$ and $-\infty+x=-\infty$ for any real number $x$, it cannot consistently be extended to define $-\infty+\infty$. Thus, even under this paradigm, saying that the sum of the limits is finite implies that the limits are finite, since the extended addition operation only yields finite numbers for finite operands.

share|cite|improve this answer

It is obvious that $\lim\limits_{n\to\infty}a_n$ and $\lim\limits_{n\to\infty}b_n$ existing and being finite implies $\lim\limits_{n\to\infty}a_n+\lim\limits_{n\to\infty}b_n$ is finite, as this is closure under addition: $x$ and $y$ existing and being finite implies $x+y$ exists and is finite.

The opposite direction depends on what you mean by $\lim\limits_{n\to\infty}a_n+\lim\limits_{n\to\infty}b_n$: would you see the assumption being met by for example $a_n=\frac{c}{2}+n$ and $b_n=\frac{c}{2}-n$? In that case it would be better to write that $\lim\limits_{n\to\infty}(a_n+b_n)$ existing and being finite does not necessarily imply $\lim\limits_{n\to\infty}a_n$ and $\lim\limits_{n\to\infty}b_n$ exist and are finite.

But I do not think that is the correct reading of $\lim\limits_{n\to\infty}a_n+\lim\limits_{n\to\infty}b_n$. Instead I think this is equivalent to the statement that $x+y$ existing and being finite implies $x$ and $y$ exist and are finite. And this I would see as true: $(+\infty)+(-\infty)$ is not meaningfully defined.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.