Evaluating Limit: n→∞ (n/(n+1)^n

[th3chemist]

Homework Statement

I'm sadly having trouble evaluating this limit.
lim n->∞ (n/(n+1)^n

Homework Equations

The Attempt at a Solution

I have not done limits for a while. I do remember you have to put it as e to the ln of it and attempt to rearrange it. Just not sure how to start.

 

[haruspex]

Try writing it as (1+1/n)^n and doing a binomial expansion.

 

[th3chemist]

Try writing it as (1+1/n)^n and doing a binomial expansion.

A binomial expansion?

 

[Mark44]

A different approach is to let y = (n/(n + 1))ⁿ and then take the ln of both sides. You can then take the limit, which will be amenable to L'Hopital's Rule (with a little manipulation first).

The idea is that under suitable conditions, the operations of taking the limit and taking the log can be interchanged.

 

[th3chemist]

A different approach is to let y = (n/(n + 1))ⁿ and then take the ln of both sides. You can then take the limit, which will be amenable to L'Hopital's Rule (with a little manipulation first).

The idea is that under suitable conditions, the operations of taking the limit and taking the log can be interchanged.

Won't that make it harder?

 

[Mark44]

Won't that make it harder?

Harder than what? The limit you have cannot be evaluated directly, as it is in the indeterminate form [1^∞].

The approach that I suggested gives you a way to evaluate this limit.

 

[th3chemist]

Harder than what? The limit you have cannot be evaluated directly, as it is in the indeterminate form [1^∞].

The approach that I suggested gives you a way to evaluate this limit.

Would I be able to write it in the form lim n-> ∞ e^ln((n/(1+n))^n)

'cause then I can bring down the n and have n*ln(n/(1+n))
then i don't know :[. Divide all by n^2?

 

[Dick]

Would I be able to write it in the form lim n-> ∞ e^ln((n/(1+n))^n)

'cause then I can bring down the n and have n*ln(n/(1+n))
then i don't know :[. Divide all by n^2?

Have you used l'Hopital's rule before? You've now got a limit of the form infinity*0. You want write in, say 0/0 form. Like ln(n/(1+n))/(1/n). Now do l'Hopital.

 

[th3chemist]

Have you used l'Hopital's rule before? You've now got a limit of the form infinity*0. You want write in, say 0/0 form. Like ln(n/(1+n))/(1/n). Now do l'Hopital.

How can you just take the ln of it though. How can you just divide by 1/n? :(

 

[Mark44]

How can you just take the ln of it though. How can you just divide by 1/n? :(

Let y = (n/(n + 1))ⁿ

Take ln of each side:
ln(y) = ln[(n/(n + 1))ⁿ]

You're not dividing by 1/n. There already is a factor of n. Multiplying by n is the same as dividing by 1/n. Since n is a very large number (and so is a long way from 0), there's no danger of division by 0 in 1/n.

Your calculus book should have one or more examples of this technique.

 

[th3chemist]

Let y = (n/(n + 1))ⁿ

Take ln of each side:
ln(y) = ln[(n/(n + 1))ⁿ]

You're not dividing by 1/n. There already is a factor of n. Multiplying by n is the same as dividing by 1/n. Since n is a very large number (and so is a long way from 0), there's no danger of division by 0 in 1/n.

Your calculus book should have one or more examples of this technique.

don't you put it the limit as e^ln (n/(n+1))^n ?

 

[Dick]

don't you put it the limit as e^ln (n/(n+1))^n ?

Do you know l'Hopital's rule or not? If you don't then this line is hopeless. Do you know lim n->infinity (1+1/n)^n=e?

 

[th3chemist]

Do you know l'Hopital's rule or not? If you don't then this line is hopeless. Do you know lim n->infinity (1+1/n)^n=e?

Yesss, I know it. I'm just trying to figure out the notation.

 

[Dick]

Yesss, I know it. I'm just trying to figure out the notation.

Ok. Then yes, once you work out lim ln((n/(n+1))^n)=c you find lim (n/(n+1))^n=e^c.

 

[ehild]

You know that $$\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$$
But

$$1+\frac{1}{n}=\frac{1+n}{n}$$

How is that related to n/(n+1)?

ehild

 

[th3chemist]

You know that $$\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n=e$$
But

$$1+\frac{1}{n}=\frac{1+n}{n}$$

How is that related to n/(n+1)?

ehild

They're the inverse of each other. So this limit would be 1/e?

 

[Mark44]

They're the inverse of each other. So this limit would be 1/e?

I would say that's a stretch. In the limit that ehild showed, the part in parentheses is (1 + n)/n. If all you do is change that part, and nothing else, it seems a remote possibility that you'll end up with the reciprocal of the limit.

There have been many good suggestions made in this thread. Why don't pick one and see what you get?

 

[th3chemist]

I would say that's a stretch. In the limit that ehild showed, the part in parentheses is (1 + n)/n. If all you do is change that part, and nothing else, it seems a remote possibility that you'll end up with the reciprocal of the limit.

There have been many good suggestions made in this thread. Why don't pick one and see what you get?

I've tried using L'Hospital's but idk what I'm doing wrong :(

 

[Dick]

I've tried using L'Hospital's but idk what I'm doing wrong :(

You haven't showed what you tried. How did you try? You can do it that way.

 

[th3chemist]

I'm doing :
Let y = (n/(n + 1))n

ln(y) = ln[(n/(n + 1))]
= n ln (n/(n+1))
= ln (n/(n+1))/(1/n)

 

[Dick]

I'm doing :
Let y = (n/(n + 1))n

ln(y) = ln[(n/(n + 1))]
= n ln (n/(n+1))
= ln (n/(n+1))/(1/n)

Ok, so you have a 0/0 form, yes? Now take d/dn of numerator and denominator. Using rules of logs will make it easier.

 

[th3chemist]

Ok, so you have a 0/0 form, yes? Now take d/dn of numerator and denominator. Using rules of logs will make it easier.

So I would get (ln(n) - ln(n+1))/(1/n)?

 

[ehild]

They're the inverse of each other. So this limit would be 1/e?

Yes. If the limit of a sequence a_n is A, and A ≠0, the limit of b_n=1/a_n is 1/A. See
http://en.wikipedia.org/wiki/Limit_of_a_sequence#Properties
or 4.3.2 at
http://science.kennesaw.edu/~plaval/math4381/seqlimthm.pdf

ehild

 

[Dick]

So I would get (ln(n) - ln(n+1))/(1/n)?

Ok, you got the log rule ok. Now l'Hopital says take the derivative of numerator and denominator, right?

 

[th3chemist]

Ok, you got the log rule ok. Now l'Hopital says take the derivative of numerator and denominator, right?

Yes sir. Which would give me (1/n - 1/(n+1))/(-1/n^2).
But then what happens next? I cannot take the limit yet.

 

[ehild]

A different approach is to let y = (n/(n + 1))ⁿ and then take the ln of both sides. You can then take the limit, which will be amenable to L'Hopital's Rule (with a little manipulation first).

The idea is that under suitable conditions, the operations of taking the limit and taking the log can be interchanged.

L'Hopital Rule is applied for functions and is derived from the laws of differentiation, those are derived from the laws of limits for functions, those are derived from the laws of limits for sequences. It is taught, and easy to derive from the definition of limit, that the limit of sum/difference of sequences is equal to the sum/difference of the limits, and if neither limit is zero/infinity, the limit of a product/ratio is equal to the product/ratio of the limits.
The limit of the sequence a_n=(1+1/n)ⁿ equal to e is very basic, the definition of the number e. The sequence in the OP is the reciprocal of a_n, so is its limit. ehild

 

[Dick]

Yes sir. Which would give me (1/n - 1/(n+1))/(-1/n^2).
But then what happens next? I cannot take the limit yet.

Use algebra to simplify the numerator. Now do you see it?

 

[Dick]

L'Hopital Rule is applied for functions and is derived from the laws of differentiation, those are derived from the laws of limits for functions, those are derived from the laws of limits for sequences. It is taught, and easy to derive from the definition of limit, that the limit of sum/difference of sequences is equal to the sum/difference of the limits, and if neither limit is zero/infinity, the limit of a product/ratio is equal to the product/ratio of the limits.
The limit of the sequence a_n=(1+1/n)ⁿ equal to e is very basic, the definition of the number e. The sequence in the OP is the reciprocal of a_n, so is its limit.


ehild

I agree that if you know lim n-> infinity (1+1/n)^n=e then there is a much faster way to solve this. But if th3chemist knows l'Hopital then working it out that way too would be a good thing.

 

[ehild]

The definition of e as a limit is taught at the beginning of the Calculus classes. Also the properties of limits.

ehild

 

[Dick]

The definition of e as a limit is taught at the beginning of the Calculus classes. Also the properties of limits.

ehild

I KNOW that. But th3chemist is pretty close to proving the limit using l'Hopital. The easy way to do it is special. l'Hopital is more general. Be a good thing to learn, yes?