MHB Calculate Limit - Is There a Special Rule?

[Petrus]

Calculate limit

I have none progress, Is there any special rule for this limit?

 

[I like Serena]

Calculate limit

I have none progress, Is there any special rule for this limit?

You can find a lower and upper bound with an integral.

 

[Petrus]

You can find a lower and upper bound with an integral.

Huh? What you mean?

 

[I like Serena]

Calculate limit

I have none progress, Is there any special rule for this limit?

You can approximate a series with an integral:
$$\sum_{k=a}^b f(k) \approx \int_{a-\frac 1 2}^{b+\frac 1 2}f(x)dx$$
More specifically, if f(x) is a strictly descending function, you have:
$$\int_{a}^{b+1}f(x)dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b}f(x)dx$$

 

[I like Serena]

Here's a picture that is somewhat indicative of how it works.

In this example the area of the rectangles is an upper bound of the integral of the function (starting from 1 in this case).
The area of the rectangles is equal to the series of the function.

 

[Petrus]

You can approximate a series with an integral:
$$\sum_{k=a}^b f(k) \approx \int_{a-\frac 1 2}^{b+\frac 1 2}f(x)dx$$
More specifically, if f(x) is a strictly descending function, you have:
$$\int_{a}^{b+1}f(x)dx < \sum_{k=a}^b f(k) < \int_{a-1}^{b}f(x)dx$$

So if I got this correct.
I will have
$$\int_{2n-\frac{1}{2}}^{6n+\frac{1}{2}}\frac{n}{k^2+n^2}dk$$ or what shall I integrate to respect?

 

[I like Serena]

So if I got this correct.
I will have
$$\int_{2n-\frac{1}{2}}^{6n+\frac{1}{2}}\frac{n}{k^2+n^2}dk$$ or what shall I integrate to respect?

Yep. That's it.

 

[Petrus]

So if I just antiderivate that It Will be $$\frac{3n}{k^3}$$ Is that correct?

 

[I like Serena]

So if I just antiderivate that It Will be $$\frac{3n}{k^3}$$ Is that correct?

Nope.
Consider what the derivative of $$\frac{3n}{k^3}$$ is with respect to k.
It's $-3 \cdot \dfrac{3n}{k^4} \ne \dfrac{n}{k^2+n^2}$

 

[Petrus]

Nope.
Consider what the derivative of $$\frac{3n}{k^3}$$ is with respect to k.
It's $-3 \cdot \dfrac{3n}{k^4} \ne \dfrac{n}{k^2+n^2}$

hmm... $$\frac{n}{k+n^2}$$ that is what I get

 

[I like Serena]

hmm... $$\frac{n}{k+n^2}$$ that is what I get

Can you show your steps starting from $$\frac{d}{dk}\left(\frac {3n}{k^3}\right)$$ then?

 

[Petrus]

Can you show your steps starting from $$\frac{d}{dk}\left(\frac {3n}{k^3}\right)$$ then?

I have problem at imagine the other as a constant and then antiderivate/derivate.. but if you ask $$\frac{3}{k^3} <=>3*\frac{1}{k^3} <=> 3*k^{-3}$$ so the derivate is $$-9k^{-4} <=> \frac{-9}{k^4}$$

 

[I like Serena]

I have problem at imagine the other as a constant and then antiderivate/derivate.. but if you ask $$\frac{3}{k^3} <=>3*\frac{1}{k^3} <=> 3*k^{-3}$$ so the derivate is $$-9k^{-4} <=> \frac{-9}{k^4}$$

Correct! :)

In your problem the integral is calculated while keeping n constant.
This means that you can treat n in just the same way as you treated the number 3.
So $$\frac{d}{dk}\left(\frac{3n}{k^3}\right) = \frac{-9n}{k^4} \ne \frac{n}{k^2+n^2}$$.So let's do this first with:
$$\int \frac{3}{x^2+3^2}dx$$
Can you find this anti-derivative?

 

[Petrus]

Correct! :)

In your problem the integral is calculated while keeping n constant.
This means that you can treat n in just the same way as you treated the number 3.
So $$\frac{d}{dk}\left(\frac{3}{k^3}\right) = \frac{-9n}{k^4} \ne \frac{n}{k^2+n^2}$$.So let's do this first with:
$$\int \frac{3}{x^2+3^2}dx$$
Can you find this anti-derivative?

Nop. But when I look in my book it looks like it will be $$\arctan$$

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Nop. But when I look in my book it looks like it will be $$\arctan$$

or Wait I think I got it now!

 

[Petrus]

Nop. But when I look in my book it looks like it will be $$\arctan$$

- - - Updated - - -or Wait I think I got it now!

$$\frac{3}{x^2+3^2}$$ if we factour out 3 from top and bot we get $$\frac{3}{3}\int\frac{1}{\frac{x^2}{3}+1^2}$$ so our integrate become $$\frac{3}{3}\arctan{\frac{x}{3}}$$ Is this correct?

 

[I like Serena]

$$\frac{3}{x^2+3^2}$$ if we factour out 3 from top and bot we get $$\frac{3}{3}\int\frac{1}{\frac{x^2}{3}+1^2}$$ so our integrate become $$\frac{3}{3}\arctan{\frac{x}{3}}$$ Is this correct?

Almost.
But your factoring out isn't quite right.
You seem to have made a mistake with the squaring part.

What do you get if you differentiate $$\frac{3}{3}\arctan{\frac{x}{3}}$$?

 

[Petrus]

Almost.
But your factoring out isn't quite right.
You seem to have made a mistake with the squaring part.

What do you get if you differentiate $$\frac{3}{3}\arctan{\frac{x}{3}}$$?

im confused now haha. 3/3=1 so we got $$\arctan{\frac{x}{3}}$$
edit: if I got this right It is wrong cause if we use chain rule we get $$\frac{1}{x^2+1}\frac{1}{3}$$ and that is not same hmm.. how I integrate that then?

 

[I like Serena]

im confused now haha. 3/3=1 so we got $$\arctan{\frac{x}{3}}$$
edit: if I got this right It is wrong cause if we use chain rule we get $$\frac{1}{x^2+1}\frac{1}{3}$$ and that is not same hmm.. how I integrate that then?

First fix the factoring out.
Your idea is correct, but you execution is not.

Then substitute u=x/3 (or equivalently x=3u).

(Btw, you did correctly find the derivative of $$\arctan{\frac{x}{3}}$$. It should give you a hint what the proper anti-derivative is.)

 

[Petrus]

First fix the factoring out.
Your idea is correct, but you execution is not.

Then substitute u=x/3 (or equivalently x=3u).

Ok now I got pretty unsure. it should be $$\frac{1/3}{1/3}$$ that I factour 1/3 from top and botom

 

[I like Serena]

Ok now I got pretty unsure. it should be $$\frac{1/3}{1/3}$$ that I factour 1/3 from top and botom

Note that $$\frac{3^2}{3} \ne 1^2$$.
And if you don't believe me, try calculating both. ;)

 

[Petrus]

Note that $$\frac{3^2}{3} \ne 1^2$$.
And if you don't believe me, try calculating both. ;)

That make sense.. I believe and can see that.. so I should have $$\frac{3}{3}\int\frac{1}{\frac{x^2}{3}+\frac{1^2}{1^2}}$$ and it will give me same result with $$\arctan{\frac{x}{3}}$$ is that correct?

 

[I like Serena]

That make sense.. I believe and can see that.. so I should have $$\frac{3}{3}\int\frac{1}{\frac{x^2}{3}+\frac{1^2}{1^2}}$$ and it will give me same result with $$\arctan{\frac{x}{3}}$$ is that correct?

No... since $$\frac{3^2}{3} \ne \frac{1^2}{1^2}$$ as well.

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You have $$\frac{3}{x^2+3^2} = \frac{3}{3}\cdot\frac{1}{\frac{x^2}{3}+\frac{3^2}{3}}$$

But what you want is something like $$\frac{1}{(\frac{x}{3})^2+1}$$

 

[Petrus]

No... since $$\frac{3^2}{3} \ne \frac{1^2}{1^2}$$ as well.

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You have $$\frac{3}{x^2+3^2} = \frac{3}{3}\cdot\frac{1}{\frac{x^2}{3}+\frac{3^2}{3}}$$

But what you want is something like $$\frac{1}{(\frac{x}{3})^2+1}$$

Do you mean
$$\frac{1}{(\frac{3x}{3})^2+1}$$

 

[I like Serena]

Do you mean
$$\frac{1}{(\frac{3x}{3})^2+1}$$

Huh? No.

I mean something like:
$$\frac{3}{x^2+3^2} = \frac{3}{3^2(\frac{x^2}{3^2}+\frac{3^2}{3^2})} = \frac{3}{3^2} \cdot \frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}$$

 

[Petrus]

Huh? No.

I mean something like:
$$\frac{3}{x^2+3^2} = \frac{3}{3^2(\frac{x^2}{3^2}+\frac{3^2}{3^2})} = \frac{3}{3^2} \cdot \frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}$$

Thanks.
So now we got $$\frac{3}{3^2}\int\frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}$$ and if we subsitute $$u=\frac{x}{3}$$ we got $$\frac{3}{3^2}\int\frac{1}{u^2+1}$$ so we got $$\frac{3}{3^2}\arctan{u}$$

 

[I like Serena]

Thanks.
So now we got $$\frac{3}{3^2}\int\frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}$$ and if we subsitute $$u=\frac{x}{3}$$ we got $$\frac{3}{3^2}\int\frac{1}{u^2+1}$$ so we got $$\frac{3}{3^2}\arctan{u}$$

Not so fast.

$$\int \frac{3}{x^2+3^2}dx = \int \frac{3}{3^2} \cdot \frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}dx
= \int \frac{1}{3} \cdot \frac{1}{(\frac{x}{3})^2+1}dx$$
We are going substitute $$u=\frac{x}{3}$$.
This means that $$du=\frac{dx}{3}$$.

Substituting gives:
$$\int \frac{1}{3} \cdot \frac{1}{(\frac{x}{3})^2+1}dx = \int \frac{1}{u^2+1}du$$

 

[Petrus]

Not so fast.

$$\int \frac{3}{x^2+3^2}dx = \int \frac{3}{3^2} \cdot \frac{1}{\frac{x^2}{3^2}+\frac{3^2}{3^2}}dx
= \int \frac{1}{3} \cdot \frac{1}{(\frac{x}{3})^2+1}dx$$
We are going substitute $$u=\frac{x}{3}$$.
This means that $$du=\frac{dx}{3}$$.

Substituting gives:
$$\int \frac{1}{3} \cdot \frac{1}{(\frac{x}{3})^2+1}dx = \int \frac{1}{u^2+1}du$$

now I integrade and put our limits?

 

[I like Serena]

now I integrade and put our limits?

I think so...

 

[Petrus]

I think so...

Hmm... when I do that and subsitute our limits to $$[\arctan{u}]_{\frac{4n-1}{6}}^{\frac{12n+1}{6}}$$ we get and then calculate limit we get $$\arctan(infinity)-arctan(infinty)$$ the answer shall be $$\arctan(6)-\arctan(2)$$ I am doing something wrong?

 

[I like Serena]

Hmm... when I do that and subsitute our limits to $$[\arctan{u}]_{\frac{4n-1}{6}}^{\frac{12n+1}{6}}$$ we get and then calculate limit we get $$\arctan(infinity)-arctan(infinty)$$ the answer shall be $$\arctan(6)-\arctan(2)$$ I am doing something wrong?

Hold on. Not so fast.
You're skipping a couple of steps.

You wanted to calculate
$$\int_{2n - \frac 1 2}^{6n + \frac 1 2} \frac{n}{k^2+n^2} dk$$

What you now have, is (note the last back substitution to x):
$$\int \frac{3}{x^2+3^2}dx = \int \frac{1}{u^2+1}du = \arctan u + C = \arctan \frac x 3 + C$$

First you need to generalize that to the actual integral that contains n.

Then you can (carefully!) substitute the boundaries.

And only then can you calculate the limit for n to infinity.

 

[Petrus]

Hold on. Not so fast.
You're skipping a couple of steps.

You wanted to calculate
$$\int_{2n - \frac 1 2}^{6n + \frac 1 2} \frac{n}{k^2+n^2} dk$$

What you now have, is (note the last back substitution to x):
$$\int \frac{3}{x^2+3^2}dx = \int \frac{1}{u^2+1}du = \arctan u + C = \arctan \frac x 3 + C$$

First you need to generalize that to the actual integral that contains n.

Then you can (carefully!) substitute the boundaries.

And only then can you calculate the limit for n to infinity.

If I got this right we got $$\arctan \frac{k}{n}+C$$

 

[Opalg]

Hmm... when I do that and subsitute our limits to $$[\arctan{u}]_{\frac{4n-1}{6}}^{\frac{12n+1}{6}}$$ we get and then calculate limit we get $$\arctan(infinity)-arctan(infinty)$$ the answer shall be $$\arctan(6)-\arctan(2)$$ I am doing something wrong?

$$\arctan(6)-\arctan(2)$$ is the correct answer. I would do this problem like this: write $$\sum_{k=2n}^{6n} \frac n{k^2+n^2}$$ as $$\sum_{k=2n}^{6n} \frac1n\,\frac 1{\bigl(\frac kn\bigr)^2+1}$$, then recognise this as a Riemann sum for the integral $$\int_2^6\frac1{x^2+1}\,dx.$$