Re: Another Sequence Problem

Quote:

Originally Posted by

**Beevo** This question asks whether the problem is converging (find limit) or diverging.

an= (n)^5n/(n-8)^5n

I first took the natural log of both the numerator and denominator and got ln(n)/ln(n-8), and then used l'hospitals rule to get a complicated mess. Can anyone give me some input on what went wrong, or if I just started out in the incorrect manner.

Thanks

I'll do the first re-arrangements to get you started ...

$\displaystyle a_n=\frac{n^{5n}}{(n-8)^{5n}} = \left(\frac{n}{n-8} \right)^{5n}~,~n> 8$

$\displaystyle \left(\frac{n}{n-8} \right)^{5n} = \left(1+\frac{8}{n-8} \right)^{5n} >\left(1+\frac{8}{n} \right)^{5n} = \left( \left(1+\frac{1}{\frac18 n} \right)^{n} \right)^5$

When calculating the limit keep in mind that $\displaystyle \lim_{t \to \infty}\left(1+\frac1t\right)^t = e$

The final result is $\displaystyle e^{40}$

Re: Another Sequence Problem

Quote:

Originally Posted by

**earboth** I'll do the first re-arrangements to get you started ...

$\displaystyle a_n=\frac{n^{5n}}{(n-8)^{5n}} = \left(\frac{n}{n-8} \right)^{5n}~,~n> 8$

$\displaystyle \left(\frac{n}{n-8} \right)^{5n} = \left(1+\frac{8}{n-8} \right)^{5n} >\left(1+\frac{8}{n} \right)^{5n} = \left( \left(1+\frac{1}{\frac18 n} \right)^{n} \right)^5$

When calculating the limit keep in mind that $\displaystyle \lim_{t \to \infty}\left(1+\frac1t\right)^t = e$

The final result is $\displaystyle e^{40}$

Thanks for the help and feedback, I really appreciate it. This is the only problem out of the sequences homework that I really couldn't understand. Actually, I am still not quite sure how you came up with your solution. Could you expand on the method more or tell me where I could find out more about it?

Thanks

Re: Another Sequence Problem

Quote:

Originally Posted by

**Beevo** Thanks for the help and feedback, I really appreciate it. This is the only problem out of the sequences homework that I really couldn't understand. **Actually, I am still not quite sure how you came up with your solution**. Could you expand on the method more or tell me where I could find out more about it?

Thanks

I don't know from your reply which part of my answer isn't understandable for you. So here is the complete version:

$\displaystyle a_n=\frac{n^{5n}}{(n-8)^{5n}} = \left(\frac{n}{n-8} \right)^{5n}~,~n> 8$ ...... Both powers have the same exponent so they could be collected to a single base.

$\displaystyle \left(\frac{n}{n-8} \right)^{5n} = \underbrace{\left(1+\frac{8}{n-8} \right)^{5n}}_{\text{result of synthetic division}} \ge \underbrace{\left(1+\frac{8}{n} \right)^{5n} }_{\text{property of a fraction}}= \underbrace{\left( \left(1+\frac{1}{\frac18 n} \right)^{n} \right)^5}_{\text{splitting the exponent}}$

By the "property of a fraction" I mean: If the denominator increases the complete fraction decreases. That explains the ">"-sign.

Since $\displaystyle \lim_{n \to \infty} \left( \frac8n \right) = \lim_{n \to \infty} \left( \frac8{n-8} \right) $ this explains the equal relation in $\displaystyle \ge$ when you are going to determine the limit.

Now calculate the limit:

$\displaystyle \lim_{n \to \infty}\left( \left(1+\frac{1}{\frac18 n} \right)^{n} \right)^5$

If $\displaystyle \frac18 n$ approaches $\displaystyle \infty$ then $\displaystyle t = \frac18 n$ approaches $\displaystyle \infty$ too.

Therefore $\displaystyle n = 8t$ . Replace n with 8t and keep in mind that $\displaystyle \lim_{t \to \infty}\left(1+\frac1t\right)^t = e$

$\displaystyle \lim_{t \to \infty}\left( \left(1+\frac{1}{\frac18 \cdot 8t} \right)^{8t} \right)^5 = \lim_{t \to \infty}\left( \left( \underbrace{\left(1+\frac{1}t \right)^{t}}_{\text{approaches e}}\right)^8 \right)^5 = \left( e^8 \right)^5 = e^{40}$

Re: Another Sequence Problem

I really have no idea, about this convergence topic, but... im trying to understand your solution step by step.... hmmm sir, may I ask, how did you use synthetic division to make (n/(n-8))^5n into (1 -(8/n-8))^5n? I really appreciate if you could explain it to me... hmmm oh yeah.. how did you use the concept of property of fraction to make 8/n-8 to 8/n? thank you very much...

Re: Another Sequence Problem

Quote:

Originally Posted by

**earboth** I don't know from your reply which part of my answer isn't understandable for you. So here is the complete version:

$\displaystyle a_n=\frac{n^{5n}}{(n-8)^{5n}} = \left(\frac{n}{n-8} \right)^{5n}~,~n> 8$ ...... Both powers have the same exponent so they could be collected to a single base.

$\displaystyle \left(\frac{n}{n-8} \right)^{5n} = \underbrace{\left(1+\frac{8}{n-8} \right)^{5n}}_{\text{result of synthetic division}} \ge \underbrace{\left(1+\frac{8}{n} \right)^{5n} }_{\text{property of a fraction}}= \underbrace{\left( \left(1+\frac{1}{\frac18 n} \right)^{n} \right)^5}_{\text{splitting the exponent}}$

By the "property of a fraction" I mean: If the denominator increases the complete fraction decreases. That explains the ">"-sign.

Since $\displaystyle \lim_{n \to \infty} \left( \frac8n \right) = \lim_{n \to \infty} \left( \frac8{n-8} \right) $ this explains the equal relation in $\displaystyle \ge$ when you are going to determine the limit.

Now calculate the limit:

$\displaystyle \lim_{n \to \infty}\left( \left(1+\frac{1}{\frac18 n} \right)^{n} \right)^5$

If $\displaystyle \frac18 n$ approaches $\displaystyle \infty$ then $\displaystyle t = \frac18 n$ approaches $\displaystyle \infty$ too.

Therefore $\displaystyle n = 8t$ . Replace n with 8t and keep in mind that $\displaystyle \lim_{t \to \infty}\left(1+\frac1t\right)^t = e$

$\displaystyle \lim_{t \to \infty}\left( \left(1+\frac{1}{\frac18 \cdot 8t} \right)^{8t} \right)^5 = \lim_{t \to \infty}\left( \left( \underbrace{\left(1+\frac{1}t \right)^{t}}_{\text{approaches e}}\right)^8 \right)^5 = \left( e^8 \right)^5 = e^{40}$

Wow! Thanks for the detailed steps for this problem. I am still puzzled as to what you did, for example the synthetic division and property of fractions. I don't ever remember learning this particular method. Is it possible to use L'hospital's rule? For example, this problem is also the same as (n/n-8)^5n, so couldn't you use the natural logarithm and then l'hospitals rule?

Re: Another Sequence Problem

Quote:

Originally Posted by

**kspkido** I really have no idea, about this convergence topic, but... im trying to understand your solution step by step.... hmmm sir, may I ask, how did you use synthetic division to make (n/(n-8))^5n into (1 -(8/n-8))^5n? I really appreciate if you could explain it to me...

There are several possible ways to do the transformation:

$\displaystyle \frac{n}{n-8} = \frac{n-8+8}{n-8} = \frac{n-8}{n-8}+\frac8{n-8} = 1+\frac{8}{n-8}$

or Code:

` 8`

n ÷ (n - 8) = 1 + -----

-(n - 8) n - 8

--------

8

This is synthetic division as it is taught in Germany.

Quote:

hmmm oh yeah.. how did you use the concept of property of fraction to make 8/n-8 to 8/n? thank you very much...

You probably have noticed that $\displaystyle n - 8 < n$. So when I changed the denominator to n **the denominator was enlarged** by 8. Thus the value of the whole **fraction became smaller**. That's all.

Re: Another Sequence Problem

Quote:

Originally Posted by

**Beevo** Wow! Thanks for the detailed steps for this problem. I am still puzzled as to what you did, for example the synthetic division and property of fractions. I don't ever remember learning this particular method. Is it possible to use L'hospital's rule? For example, this problem is also the same as (n/n-8)^5n, so couldn't you use the natural logarithm and then l'hospitals rule?

This is basic to everything done here: $\displaystyle {\lim _{n \to \infty }}{\left( {1 + \frac{1}{n}} \right)^n} = e$

That can be generalized to: $\displaystyle {\lim _{n \to \infty }}{\left( {1 + \frac{a}{n+b}} \right)^{cn}} = e^{ac}$

So here is an easy example: $\displaystyle {\lim _{n \to \infty }}{\left( {\frac{{n - 3}}{{n + 1}}} \right)^{2n}} = {\lim _{n \to \infty }}{\left( {1 + \frac{{ - 4}}{{n + 1}}} \right)^{2n}} = {e^{ - 8}}$

Re: Another Sequence Problem

Got it... I'll try and memorize this method. Thanks

Re: Another Sequence Problem

Quote:

Originally Posted by

**Beevo** This question asks whether the problem is converging (find limit) or diverging.

an= (n)^5n/(n-8)^5n

I first took the natural log of both the numerator and denominator and got ln(n)/ln(n-8), and then used l'hospitals rule to get a complicated mess. Can anyone give me some input on what went wrong, or if I just started out in the incorrect manner.

Thanks

An easier method:

$\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left[\frac{n^{5n}}{(n - 8)^{5n}}\right] &= \lim_{n \to \infty}\left(\frac{n}{n - 8}\right)^{5n} \\ &= \lim_{n \to \infty}e^{\ln{\left[ \left( \frac{n}{n - 8} \right)^{5n} \right]}} \\ &= \lim_{n \to \infty}e^{5n\ln{\left(\frac{n}{n-8}\right)}} \\ &= e^{\lim_{n \to \infty}5n\ln{\left( \frac{n}{n-8} \right)}} \\ &= e^{\lim_{n \to \infty}5\left[ \frac{\ln{(n)} - \ln{(n-8)}}{\frac{1}{n}} \right]} \\ &= e^{\lim_{n \to \infty} 5\left( \frac{ \frac{1}{n} - \frac{1}{n - 8} }{ -\frac{1}{n^2} } \right) } \textrm{ by L'Hospital's Rule} \\ &= e^{\lim_{n \to \infty} 5 \left[ \frac{-\frac{8}{n(n-8)}}{ -\frac{1}{n^2} } \right]} \\ &= e^{\lim_{n \to \infty} 40\left( \frac{n^2}{n^2 - 8n} \right)} \\ &= e^{\lim_{n \to \infty} 40 \left( \frac{1}{1 - \frac{8}{n}} \right) } \\ &= e^{40} \end{align*}$

Re: Another Sequence Problem

Quote:

Originally Posted by

**Prove It** An easier method:

$\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left[\frac{n^{5n}}{(n - 8)^{5n}}\right] &= \lim_{n \to \infty}\left(\frac{n}{n - 8}\right)^{5n} \\ &= \lim_{n \to \infty}e^{\ln{\left[ \left( \frac{n}{n - 8} \right)^{5n} \right]}} \\ &= \lim_{n \to \infty}e^{5n\ln{\left(\frac{n}{n-8}\right)}} \\ &= e^{\lim_{n \to \infty}5n\ln{\left( \frac{n}{n-8} \right)}} \\ &= e^{\lim_{n \to \infty}5\left[ \frac{\ln{(n)} - \ln{(n-8)}}{\frac{1}{n}} \right]} \\ &= e^{\lim_{n \to \infty} 5\left( \frac{ \frac{1}{n} - \frac{1}{n - 8} }{ -\frac{1}{n^2} } \right) } \textrm{ by L'Hospital's Rule} \\ &= e^{\lim_{n \to \infty} 5 \left[ \frac{-\frac{8}{n(n-8)}}{ -\frac{1}{n^2} } \right]} \\ &= e^{\lim_{n \to \infty} 40\left( \frac{n^2}{n^2 - 8n} \right)} \\ &= e^{\lim_{n \to \infty} 40 \left( \frac{1}{1 - \frac{8}{n}} \right) } \\ &= e^{40} \end{align*}$

Thanks for this detailed response.