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This is a real-analysis homework question so I of course have to be very precise and justify anything or any theorem I use.

If $b_n$ is a bounded sequence and $\lim(a_n) = 0$, show that $\lim(a_nb_n) = 0$

Intuitively, since $b_n$ is bounded, then sup($b_n$) is some finite number and therefore we can take an $N$ natural number as large as we need such that for all $n\gt N$ $b_na_n$ approaches $0$.

At first I thought to use the limit theorems, but since $a_n$ is not bounded, the general limit theorems do not reply. (I am referring to $\lim(X + Y) = \lim X + \lim Y$ for $X,Y$ sequences etc).

I was thinking then to use the definition of the limit somehow to show that since $b_n$ is bounded we can take as intuitively stated above $N$ large enough to show the statement is true. I'm not sure how to proceed with this.

Thank you for your replies in advance!

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    $\begingroup$ Prove first that $(a_n)$ is bounded, if you need it. $\endgroup$ – Mariano Suárez-Álvarez Nov 10 '10 at 4:50
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You want to show that the absolute value of $b_na_n$ can be made arbitrarily small. But, there exists a positive constant $M$ such that for all $n$, $|b_n|\leq M$, since the sequence $(b_n)$ is bounded. Thus $|b_na_n|\leq M|a_n|$ for all $n$. Since the sequence $(a_n)$ converges to 0, this is all you should need. The argument is basically the Squeeze Theorem from elementary calculus.

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  • $\begingroup$ Note that we have to use a couple theorems here to show this since we are being very precise. We use that $|a_n||b_n| = |a_nb_n|$ and also that $Theorem$ If $a_n$ is a convergent sequence, then $a_n$ is bounded. We also have to use limit theorems to show that $lim(C|a_n|) = Clim|a_n|$ Lastly, we need the theorem that states that if $|a_n| < C$ for some $C>0$ real number, then $-C < a_n < C$ $\endgroup$ – Justin Nov 11 '10 at 17:34
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Since the $b_n$ are bounded, there is a constant $C$ such that $-C\lt b_n\lt C$ for all $n$. Fix $n$. Suppose first $a_n\gt 0$. Then $-C a_n< a_nb_n < Ca_n$. Similarly, if $a_n\lt 0$ then $Ca_n < a_n b_n < -C a_n$. In short: $|a_n b_n|\lt C|a_n|$.

Now, for any sequence $c_n$, $c_n\to 0$ iff $|c_n|\to 0$. Right?

But $C|a_n|\to 0$ since $a_n\to 0$. Since $|a_nb_n|\le C|a_n|$, we must also have $|a_n b_n|\to 0$.

(This is just a sketch. Let me know if you need me to clarify something.)

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You write *... but since $a_n$ is not bounded...*

Presumably you mean that you have not been explicitly told that the sequence $a_n$ is bounded. That is true; you have not.

However, as it happens, the following holds:

Theorem. Let $c_n$ be any sequence of real numbers. If $c_n$ converges, then it is bounded.

Intuitively, why is this true? Well, for large enough $n$, the values of $c_n$ must be close to the limit $L$; so only the "early" values of $c_n$ can get away from the limit, and since there are only finitely many of them, they can only go so far.

So, try to prove this carefully, so you can conclude that $a_n$ is bounded, so that the theorems you want to apply can be applied.

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    $\begingroup$ @Chandru1: meaning... what? $\endgroup$ – Arturo Magidin Nov 10 '10 at 5:12
  • $\begingroup$ @Dan petersen: Since it doesn't make sense i have deleted it. $\endgroup$ – anonymous Nov 10 '10 at 12:06
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I think this old question deserves an answer straight from the definition of a limit.

The sequence $(b_n)$ is bounded, take an $M$ such that $\left|b_n\right|<M$ for all $n$.

Now let $\varepsilon>0$. Since $(a_n)$ converges to $0$ there is an $N$ such that $|a_n|<\frac\varepsilon M$ for all $n\ge N$. Then $|a_n b_n|<\frac\varepsilon M\cdot M=\varepsilon$ for all $n\ge N$. This shows that $(a_nb_n)$ converges to $0$.

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HINT: Let $N(\epsilon)$ be such that $\frac{1}{N(\epsilon)}<\frac{\epsilon}{M}$, where $|b_{n}| \leq M$ ( since $b_{n}$ is bounded.) Then think of $|a_{n}b_{n} - 0|$ for $n \geq N(\epsilon)$.

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