# Proving {$b_n$}$_{n=1}^\infty$ converges given {$a_n$}$_{n=1}^\infty$ and {$a_n b_n$}$_{n=1}^\infty$

Suppose {$a_n$}$_{n=1}^\infty$ and {$b_n$}$_{n=1}^\infty$ are sequences such that {$a_n$}$_{n=1}^\infty$ coverges to A$\neq$0 and {$a_n b_n$}$_{n=1}^\infty$ converges. Prove that {$b_n$}$_{n=1}^\infty$ converges.

What I have so far:

$b_n = {a_n b_n \over a_n}$ $\to$ $C \over A$, $A\neq0$

|$b_n - {C \over A}$| = |${a_n b_n \over a_n} - {C \over A}$| = |${Aa_nb_n - Ca_n \over Aa_n}$| $\leq$ |${1 \over Aa_n}||Aa_nb_n - Ca_n$|=|${1 \over Aa_n}||a_n(Ab_n - C)$|

$\leq |{1 \over Aa_n}||a_n||(Ab_n - C)$|. Note: since $a_n$ converges, there is M>0 such that |$a_n| \leq$M for all n $\in\Bbb N$.

Thus, |${1 \over Aa_n}||a_n||(Ab_n - C)$| = |${1 \over M}||M||(Ab_n - C)$|. And this is where I get lost. Any thoughts? Or am I completely wrong to begin with?

• well the problem is that $a_n$ could be $0$ for several (finitely) $n$. The absolute value is multiplicative by the way so you have = everywhere – Dominic Michaelis Feb 19 '13 at 16:11
• It is possible (likely even) that you are supposed to show that $1/a_n\to 1/A$, and then apply the result that tells something nice about the convergence of the product of two converging sequences. – Jyrki Lahtonen Feb 19 '13 at 16:39
• Alternatively (getting your hands dirty) write $$\frac{Aa_nb_n-Ca_n}{Aa_n}=\frac{A(a_nb_n-C)+C(A-a_n)}{Aa_n}$$ and then use the facts that 1) $|a_nb_n-C|$ is small, when $n$ is large, 2) $|A-a_n|$ is small, when $n$ is large, 3) $1/|Aa_n|$ is bounded, when $n$ is large. – Jyrki Lahtonen Feb 19 '13 at 16:43

Mh what is about something like:
$a_n b_n$ converges now let's call the limit $C$, as $A\neq 0$ we can write $C=A\cdot B$ with $B=\frac{C}{A}$ \begin{align*} |a_n b_n - A \cdot B| &= |a_n b_n - b_n A +b_n A - A\cdot B|\\ &=|b_n(a_n-A) + A(b_n-B)| \\ \end{align*} Because $a_n$ converges the first part converges to zero, as we know the lhs converges the rhs has to converges to, so $$|A(b_n-B)|=|A| |b_n-B|$$ must converge to zero, as we know $|A|\neq 0$ we know $$|b_n-B|$$ converges to $0$. And so $b_n$ converges to $B$

As robjohn pointed out we get the triangle inequality $$|A(b_n-B)|\leq |a_n b_n - AB| + |b_n(a_n-A)|$$ As we know $a_nb_n$ converges with $a_n$ not converging to $0$. If $b_n$ is not bounded $a_n b_n$ can't converge, as we know $a_nb_n$ is convergent, we get $b_n$ is bounded.

• What is $B$? You cannot assume the existence of the limit while you are trying to prove the existence of the limit. – robjohn Feb 19 '13 at 17:28
• oh I forgot to say it, thanks for pointing out. – Dominic Michaelis Feb 19 '13 at 17:31
• You get the triangle inequality $$|A(b_n-B)|\le|a_nb_n-AB|+|b_n(a_n-A)|$$ we know that $|a_nb_n-AB|\to0$ and $|a_n-A|\to0$. Do we have any control of $b_n$ on the RHS? – robjohn Feb 19 '13 at 17:38
• Thank you both this helps a lot. – Student Feb 19 '13 at 17:57
• @DominicMichaelis: Your argument has the right spirit, but I still feel it lacks rigor. When you say that $a_nb_n$ converges with $a_n$ not converging to $0$, what you should be saying is: $a_n$ converges to a non-zero limit. To justify that this shows $b_n$ is bounded, the argument should refer to an inequality with $b_n$ on one side and not on the other, such as: since $a_n\to A$ and $a_nb_n\to AB$ then for big enough $n$, $|a_n|\ge\frac{|A|}{2}$ and $|a_nb_n|\le2|AB|$, thus $|b_n|=\frac{|a_nb_n|}{|a_n|}\le\frac{2|AB|}{\frac12|A|}=4|B|$. – robjohn Feb 19 '13 at 20:58

Hint: Let $A=\lim\limits_{n\to\infty}a_n$ and $B=\frac{\lim\limits_{n\to\infty}a_nb_n}{A}$.

Since $|A|>0$, for $n$ large enough, $|a_n-A|\le\frac{|A|}{2}$. Show that then, $|a_n|\ge\frac{|A|}{2}$.

Then note that $$a_n(b_n-B)=(a_nb_n-AB)-B(a_n-A)$$

$$\frac{\lim_{n\to\infty}a_n b_n}{\lim_{n\to\infty}a_n} = \lim_{n\to\infty}\frac{a_n b_n}{a_n} =\lim_{n\to\infty}b_n.$$

This is legitimate because you start with limits that you know exist. The rest is standard properties of limits of sequences found in any calc textbook.