Evaluating Limit: n→∞ (n/(n+1)^n

[th3chemist]

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

Would I get -n + n^2/(n+1)
in which I would then divide everything by n to get -1 + n/(n+1)?

 

[Dick]

Would I get -n + n^2/(n+1)
in which I would then divide everything by n to get -1 + n/(n+1)?

No. Why don't you show how you got for a change, instead of making us guess?

 

[th3chemist]

No. Why don't you show how you got for a change, instead of making us guess?

Don't you just multiply by the denominator?

 

[Mark44]

Don't you just multiply by the denominator?

Show us what you did.

 

[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?

Well, L'Hopital can be applied formally, but in case the sequence is called "function" it is defined for positive integers only. It has no derivative. You should apply L'Hopital rule to the function (x/(1+x))^x, and then refer to the theorem that the limit of all subsets of x tends to the limit of the function. Neither is it straightforward that the limit of logarithm is equal to the logarithm of the limit. You need to be strict in Maths.

ehild

 

[th3chemist]

Show us what you did.

((1/n) -(1/(n+1))(-n^-2)

Rats. I think this is where I made a mistake. I take the derivative of 1/n right? to get -1/n^2.
which gives -1/n^3 + 1/(n^2(n+1))

Though I feel like this is wrong -_-

 

[Mark44]

You're doing OK. Simplify the first part - ((1/n) -(1/(n+1)) - by combining the fractions.

 

[th3chemist]

You're doing OK. Simplify the first part - ((1/n) -(1/(n+1)) - by combining the fractions.

Ah.

That gives you 1/(n(n+1)). Now I multiply the -n^-2 to get 1/(n^3(n+1))?

 

[Mark44]

((1/n) -(1/(n+1))(-n^-2)

You have a sign wrong at the end, which I missed before.

You're dividing by -1/n², so invert this and multiply, which gives you
((1/n) -(1/(n+1))(-n^+2)

Rats. I think this is where I made a mistake. I take the derivative of 1/n right? to get -1/n^2.
which gives -1/n^3 + 1/(n^2(n+1))

Though I feel like this is wrong -_-

 

[th3chemist]

You have a sign wrong at the end, which I missed before.

You're dividing by -1/n², so invert this and multiply, which gives you
((1/n) -(1/(n+1))(-n^+2)

But isn't the derivative of 1/n -1/n^2 ? why would the sign be positive?

If I multiply -n^2 into the equation I get -n + n^2/(n+1)
I presume I add the fractions to get (-n^2 -n + n^2)/(n+1) = -n/(n+1).

 

[Mark44]

But isn't the derivative of 1/n -1/n^2 ? why would the sign be positive?

Yes, d/dn(1/n) = -1/n²
I wrote this as -1/n⁺² because in your work, you had a negative sign on the exponent.

If I multiply -n^2 into the equation I get -n + n^2/(n+1)
I presume I add the fractions to get (-n^2 -n + n^2)/(n+1) = -n/(n+1).

Yes

 

[th3chemist]

Yes, d/dn(1/n) = -1/n²
I wrote this as -1/n⁺² because in your work, you had a negative sign on the exponent.

Yes

1/n^2 = -n^-2 though. Oh wait I think I see it now.

What can I do after I have -n/(n+1)? I can't take the limit yet. It would be ∞/∞

 

[Mark44]

We are 42 posts into this, so you might not be keeping track of what you're trying to do, so let's summarize.

The original problem is to evaluate this limit:
$$ \lim_{n \to \infty} \left(\frac{n}{n+1} \right)^n$$

The track you're taking was to let y = (n/(n + 1))ⁿ

The next step was to take the natural log of both sides, leading to
ln(y) = n ln[n/(n + 1)]

You then took the limit of both sides. See if you can write the equation that represents this.

 

[th3chemist]

We are 42 posts into this, so you might not be keeping track of what you're trying to do, so let's summarize.

The original problem is to evaluate this limit:
$$ \lim_{n \to \infty} \left(\frac{n}{n+1} \right)^n$$

The track you're taking was to let y = (n/(n + 1))ⁿ

The next step was to take the natural log of both sides, leading to
ln(y) = n ln[n/(n + 1)]

You then took the limit of both sides. See if you can write the equation that represents this.

I just don't see it :(.

And the limit for -n/(n+1) = -1. As you divide n by the top and bottom. So the answer should be e^-1?

 

[Mark44]

We are 42 posts into this, so you might not be keeping track of what you're trying to do, so let's summarize.

The original problem is to evaluate this limit:
$$ \lim_{n \to \infty} \left(\frac{n}{n+1} \right)^n$$

The track you're taking was to let y = (n/(n + 1))ⁿ

The next step was to take the natural log of both sides, leading to
ln(y) = n ln[n/(n + 1)]

You then took the limit of both sides. See if you can write the equation that represents this.

And the limit for -n/(n+1) = -1. As you divide n by the top and bottom. So the answer should be e^-1?

See if you can write the equations that represent what I summarized above.

 

[th3chemist]

See if you can write the equations that represent what I summarized above.

So my answer is wrong? :(

 

[Mark44]

Did I say that?

I'm trying to get you to write a coherent, logical sequence of mathematical statements.