# Limits

• Jan 24th 2008, 02:08 PM
Joel24
Limits
Hey, I'm stuck on this question and desperately need help.

Question

Suppose that the sequence $(a_n)$ is convergent, and assume that the sequence $(b_n)$ is bounded. Prove that the sequence $(c_n)$ defined by

$c_n = \frac{a_nb_n - 5n}{a^2_n + 3n}$

is convergent and find its limit. (n belongs to the set of natural numbers)

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Please help! I tried using the theorem that if a sequence is convergent, then it is bounded (for $(a_n)$) but get anywhere. :( It's quite clear to me that the limit is $\frac{-5}{3}$ but I don't know how to get there.
• Jan 24th 2008, 04:49 PM
ThePerfectHacker
Quote:

Originally Posted by Joel24
Hey, I'm stuck on this question and desperately need help.

Question

Suppose that the sequence $(a_n)$ is convergent, and assume that the sequence $(b_n)$ is bounded. Prove that the sequence $(c_n)$ defined by

$c_n = \frac{a_nb_n - 5n}{a^2_n + 3n}$

Since $c_n$ has a possibility is not being properly defined we should make a restriction like $a_n,b_n\geq 0$.

Hint: Divide the numerator and denominator by $n$ and take the limit.
• Jan 25th 2008, 05:32 AM
Joel24
Quote:

Originally Posted by ThePerfectHacker
Since $c_n$ has a possibility is not being properly defined we should make a restriction like $a_n,b_n\geq 0$.

Hint: Divide the numerator and denominator by $n$ and take the limit.

Ok, thanks for the reply. Here's what I've done:
__________________________________________________ _____________

$c_n = \frac{ \frac{a_n b_n}{n} - 5}{\frac{a^2_n}{n} + 3}$

Let $x \in \{ a_n : n$ is a natural number $\}$, and let $M_1$ be an upper bound of the set.

Let $y \in \{ b_n : n$ is a natural number $\}$, and let $M_2$ be an upper bound of the set.

Then, for all $a_n, b_n > 0$

$\lim_{n \to \infty} \left( \frac{ \frac{xy}{n} - 5}{ \frac{x^2}{n} + 3} \right)$

But as $x \leq M_1$ for all $x$ and y $y \leq M_2$ for all $y$, it follows that

$\lim_{n \to \infty} \left( \frac{xy \left( \frac{1}{n}\right) - 5}{x^2 \left( \frac{1}{n} \right) + 3} \right) = \frac{-5}{3}$

__________________________________________________ ______________

But I don't know how to prove the sequence is convergent before finding the limit as the question asks.

Could I perhaps show the sequence is monotone and bounded and therefore convergent? I've tried, but couldn't prove it.

Also, what if $a_n < 0$ and $b_n \geq 0$? We made a restriction but surely this doesn't cover all possibilities.

Sorry if I'm asking silly questions, but I really need help with this. :o
• Jan 25th 2008, 09:23 AM
ThePerfectHacker
Quote:

Originally Posted by Joel24
Also, what if $a_n < 0$ and $b_n \geq 0$? We made a restriction but surely this doesn't cover all possibilities.

Sorry if I'm asking silly questions, but I really need help with this.

If we do not make this restrictions then the denominator can be zero and the sequences are not well-defined. It is absolutely necessary to make some sort of restriction.

Just use the fact that is $a_n$ is bounded then $\frac{a_n}{n} \to 0$. Thus, if a sequence $a_n$ is convergent (therefore bounded) we have $\frac{a_n}{n}\to 0$.

This means,
$\frac{\frac{a_nb_n}{n} - 5}{\frac{a_n^2}{n} + 3} \to -\frac{5}{3}$

Because $a_nb_n$ is bounded and $a_n^2$ is bounded.
• Jan 25th 2008, 10:36 AM
Joel24
Ok, thank you. :)

In the seminars, we haven't been told that $lim_{n \to \infty} \left(\frac{a_n}{n} \right) = 0$ if the sequence is bounded. As obvious as it is. :p

So do you think it would be more 'rigorous' if I said that since $a_n$ is bounded, there exists an $M$ such that $| a_n | \leq M$

Then I can show that $lim_{n \to \infty} \left(\frac{M}{n} \right) = 0$ and $lim_{n \to \infty} \left(\frac{-M}{n} \right) = 0$. Since $-M \leq a_n \leq M$, $\forall n$ then by the sandwich theorem, $lim_{n \to \infty} \left(\frac{a_n}{n} \right) = 0$

Is that fine, in your opinion, or unnecessary?

Thanks.
• Jan 25th 2008, 10:48 AM
ThePerfectHacker
Quote:

Originally Posted by Joel24
Ok, thank you. :)

In the seminars, we haven't been told that $lim_{n \to \infty} \left(\frac{a_n}{n} \right) = 0$ if the sequence is bounded. As obvious as it is. :p

So do you think it would be more 'rigorous' if I said that since $a_n$ is bounded, there exists an $M$ such that $| a_n | \leq M$

Then I can show that $lim_{n \to \infty} \left(\frac{M}{n} \right) = 0$ and $lim_{n \to \infty} \left(\frac{-M}{n} \right) = 0$. Since $-M \leq a_n \leq M$, $\forall n$ then by the sandwich theorem, $lim_{n \to \infty} \left(\frac{a_n}{n} \right) = 0$
That is excellent.
Is that fine, in your opinion, or unnecessary?

Thanks.

That is excellent.