# If $a_nb_n\to 1$ and $a_n+b_n\to2$, do $a_n,b_n\to1$?

Suppose that $a_n$ and $b_n$ are two real sequences satisfying $a_nb_n\to 1$ and $a_n+b_n\to 2$. Does it follow that $a_n$ and $b_n$ both converge to $1$?

I was working on this problem as one of the exercises from this online source of problems in analysis. I have not been able to find a counterexample, but I have deduced some necessary conditions on $a_n,b_n$:

1. Eventually $a_n$ and $b_n$ have the same sign, and this must be $>0$:

If they do not eventually share the same sign, then $a_nb_n$ can't converge to $1$. If they are not both positive, their sum can't converge to $2$.

2. There exists $t>0$ and $N$ such that $a_n>t$ for all $n\ge N$. [Hence the same conclusion holds for $b_n$.]:

If not, there is some subsequence $a_{n_k}$ with $a_{n_k}\to 0$. But then $b_{n_k}\to\infty$ as $a_{n_k}b_{n_k}\to1$. Then $a_{n_k}+b_{n_k}\to\infty$, absurd.

3. $a_n$ and $b_n$ are both bounded above:

Otherwise, there must be a subsequence of the other that tends to $0$, which is impossible by 2.

I imagine a counterexample would be very strange, but I feel like a proof is also not too far out of reach.

Thanks for any suggestions.

P.S. Does the conclusion change if we allow complex sequences $a_n,b_n$?

• $(a_n - b_n)^2 = (a_n + b_n)^2 - 4a_nb_n \to 2^2 - 4\cdot 1 = 0$. It follows that $a_n - b_n \to 0$, and then $a_n = \frac{(a_n + b_n) + (a_n - b_n)}{2} \to \frac{2+0}{2} = 1$. What one needs is the implication $c_n^2 \to 0 \implies c_n \to 0$ (and limit arithmetic, and one must be able to divide by $2$, so the argument doesn't work in characteristic $2$, but it works in any commutative ring with its topology induced by an absolute value in which $2$ is a unit). – Daniel Fischer Aug 30 '18 at 20:30

Hint. $|a_n - b_n| = \sqrt{(a_n+b_n)^2-4a_n b_n}$ converges to zero. From this, both

$$\max\{a_n, b_n\} = \frac{(a_n+b_n)+|a_n-b_n|}{2}, \qquad \min\{a_n, b_n\} = \frac{(a_n+b_n)-|a_n-b_n|}{2}$$

converge to $1$. Can you now finish the proof by applying the squeezing lemma?

The same conclusion holds for complex case but we may need a slightly more lengthy solution. First, we still know that

$$|a_n - b_n| = |(a_n+b_n)^2-4a_n b_n|^{1/2}$$

converges to $0$. Then by the triangle inequality,

$$|a_n|, |b_n| \leq \frac{|a_n+b_n|+|a_n-b_n|}{2}$$

and hence both $(a_n)$ and $(b_n)$ are bounded. Now for any subsequene of $(a_n, b_n)$, we can extract a further subsequence, say $(a_{n_k}, b_{n_k})$ such that both components converge. Then by the relation $a_{n_k} + b_{n_k} \to 2$ and $|a_{n_k} -b_{n_k}|\to 0$, it follows that both $(a_{n_k})$ and $(b_{n_k})$ converge to $1$. This implies that the original sequence $(a_n, b_n)$ also converges to $(1,1)$ by the following well-known trick:

Lemma. For a metric space $(X, d)$ and for a sequence $(x_n)$ in $M$, the followings are equivalent:

1. $(x_n)$ converges to $x\in M$.

2. Any subsequence of $(x_n)$ has a further subsequence that converges to $x$.

• Thanks very much! Yes, I can take it from here. – Alex Ortiz Jul 9 '17 at 6:33

Consider $$c_n=(a_n-1)^2+(b_n-1)^2.$$ Then $$c_n=(a_n+b_n)^2-2a_nb_n-2(a_n+b_n)+2\to2^2-2-4+2=0.$$ As $0\le(a_n-1)^2\le c_n$ then $a_n-1\to0$, and $a_n\to1$. Similarly $b_n\to1$.

• Nice solution(+1)! Do you expect a similar, nice algebraic trick for complex case too? – Sangchul Lee Jul 9 '17 at 6:52

Alternatively, the polynomials $p_n(x) = (x-a_n)(x-b_n)$ will converge pointwise to the function $p(x) = (x-1)^2.$ It should not be difficult to take it from here, since they are all polynomials.

• I like this method of proof, but could you perhaps spell it out a little further how exactly knowing that $p_n\to p$ implies that its roots converge to $1$? – Alex Ortiz Jul 9 '17 at 16:00

Let $s_n := a_n + b_n$, $p_n := a_n b_n$. Then by the quadratic formula, $$a_n, b_n = \frac{s_n \pm \sqrt{s_n^2 - 4 p_n}}{2}.$$ Now, by the hypotheses, $s_n \to 2$ and $p_n \to 1$ as $n \to \infty$, so $s_n^2 - 4 p_n \to 0$. Therefore, $\sqrt{s_n^2 - 4 p_n} \to 0$ also, and from here it should be easy to prove that $a_n, b_n \to 1$.

(This proof should also be valid, with no changes, in the case where $a_n, b_n$ are complex sequences - since whatever branch cut you choose for the function $\sqrt{z}$, it's still continuous at 0.)