Evaluating Limit: n→∞ (n/(n+1)^n

[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.