# Thread: Convergence of a sequence example

1. ## Convergence of a sequence example

Starting to look at Convergence of sequences and given an example without a solution. The solution would really help my understanding of what's going on here:

$\displaystyle x_{n} = \frac{9n^{3}+3}{2n^{3}+5}$

Define what is meant by saying $\displaystyle x_{n}$ converges to the limit l as n goes to infinity. Show that $\displaystyle x_{n}$ converges to $\displaystyle \frac{9}{2}$ as n tends to infinity.

Now I know you can pull out the fraction and show that the rest tends to 1 (right?), simple enough you can already see the $\displaystyle \frac{9}{2}$ right there but how do you show this? Basically that l = $\displaystyle \frac{9}{2}$

2. ## Re: Convergence of a sequence example

Perhaps you got typo, as written, the sequence diverges.

3. ## Re: Convergence of a sequence example

Originally Posted by Shizaru
Starting to look at Convergence of sequences and given an example without a solution. The solution would really help my understanding of what's going on here:
$\displaystyle x_{n} = \frac{9n^{3}+3}{2n^{2}+5}$
Define what is meant by saying $\displaystyle x_{n}$ converges to the limit l as n goes to infinity. Show that $\displaystyle x_{n}$ converges to $\displaystyle \frac{9}{2}$ as n tends to infinity.
There is a real problem with your post:
$\displaystyle \lim _{n \to \infty } \frac{9n^{3}+3}{2n^{2}+5} = \infty$

If it were $\displaystyle \lim _{n \to \infty } \frac{9n^{2}+3}{2n^{2}+5} = \frac{9}{2}$ that is correct.

Which is it?

4. ## Re: Convergence of a sequence example

Yep, sorry, both powers are 3. Corrected it.

5. ## Re: Convergence of a sequence example

Originally Posted by Shizaru
Starting to look at Convergence of sequences and given an example without a solution. The solution would really help my understanding of what's going on here:

$\displaystyle x_{n} = \frac{9n^{3}+3}{2n^{3}+5}$

Define what is meant by saying $\displaystyle x_{n}$ converges to the limit l as n goes to infinity. Show that $\displaystyle x_{n}$ converges to $\displaystyle \frac{9}{2}$ as n tends to infinity.

Now I know you can pull out the fraction and show that the rest tends to 1 (right?), simple enough you can already see the $\displaystyle \frac{9}{2}$ right there but how do you show this? Basically that l = $\displaystyle \frac{9}{2}$
The standard way to deal with a problem like this is to divide top and bottom of the fraction by the highest power of n in sight, and then use some theorems about limits of sums, products and quotients. In this case, the first step is to divide through by $\displaystyle n^3$:

$\displaystyle \frac{9n^{3}+3}{2n^{3}+5} = \frac{9+\frac3{n^3}}{2+\frac5{n^3}}.$

In the numerator, $\displaystyle \tfrac3{n^3}\to0$ as $\displaystyle n\to\infty.$ There is a theorem saying that the limit of a sum is the sum of the limits. That theorem tells you that the limit of the numerator is 9+0=9. Similarly, the limit of the denominator is 2+0=2. Finally, another theorem says that if the numerator and denominator of a fraction both have limits (and the denominator has a nonzero limit) then the limit of the quotient is the quotient of the limits. Apply that theorem to see that the limit of $\displaystyle \frac{9+\frac3{n^3}}{2+\frac5{n^3}}$ is 9/2.

6. ## Re: Convergence of a sequence example

Another way (using the definition of limit):

$\displaystyle \left |\frac{9n^3+3}{2n^3+5}-\frac{9}{2}\right |<\epsilon \Leftrightarrow \frac{39}{4n^3+10}<\epsilon \Leftrightarrow \sqrt[3]{\frac{39-10\epsilon}{4\epsilon}}<n$

This condition is satisfied for $\displaystyle n\geq n_0=\left\lfloor \sqrt[3]{\frac{39-10\epsilon}{4\epsilon}}\right\rfloor +1$

7. ## Re: Convergence of a sequence example

Thanks - how would I write up something like this. I know the theorems, and that basically each part can be taken on its own, like as you put it sum of limits = limit of sum and likewise for quotient. I know these rules but I don't know how to express it... to me it just seems like proving 1 + 1 = 2, its intuitive but I need a way of expressing it... I hope you get what I mean,,

Plus what I was trying to say is that I would rearrange $\displaystyle x_{n}$ into ($\displaystyle \frac{9}{2}$)($\displaystyle \frac{n^{3} + \frac{1}{3}}{n^{3}+\frac{5}{2}}$) show that the second bracket tends to 1 .. is this the same as what you were saying? (just the opposite?) this was my initial working.