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
\(\displaystyle \int_{2n-\frac{1}{2}}^{6n+\frac{1}{2}}\frac{n}{k^2+n^2}dk\) or what shall I integrate to respect?

 

[I like Serena]

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

Yep. That's it.

 

[Petrus]

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

 

[I like Serena]

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

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

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

 

[I like Serena]

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

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

 

[Petrus]

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

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

 

[I like Serena]

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

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 \(\displaystyle \frac{d}{dk}\left(\frac{3n}{k^3}\right) = \frac{-9n}{k^4} \ne \frac{n}{k^2+n^2}\).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 \(\displaystyle \frac{d}{dk}\left(\frac{3}{k^3}\right) = \frac{-9n}{k^4} \ne \frac{n}{k^2+n^2}\).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 \(\displaystyle \arctan\)

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

or Wait I think I got it now!

 

[Petrus]

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

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

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

 

[I like Serena]

\(\displaystyle \frac{3}{x^2+3^2}\) if we factour out 3 from top and bot we get \(\displaystyle \frac{3}{3}\int\frac{1}{\frac{x^2}{3}+1^2}\) so our integrate become \(\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}\) 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 \(\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}\)?

 

[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 \(\displaystyle \frac{3}{3}\arctan{\frac{x}{3}}\)?

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

 

[I like Serena]

im confused now haha. 3/3=1 so we got \(\displaystyle \arctan{\frac{x}{3}}\)
edit: if I got this right It is wrong cause if we use chain rule we get \(\displaystyle \frac{1}{x^2+1}\frac{1}{3}\) 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 \(\displaystyle \arctan{\frac{x}{3}}\). 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 \(\displaystyle \frac{1/3}{1/3}\) that I factour 1/3 from top and botom

 

[I like Serena]

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

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

 

[Petrus]

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

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

 

[I like Serena]

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

No... since \(\displaystyle \frac{3^2}{3} \ne \frac{1^2}{1^2}\) as well.

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You have \(\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3}\cdot\frac{1}{\frac{x^2}{3}+\frac{3^2}{3}}\)

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

 

[Petrus]

No... since \(\displaystyle \frac{3^2}{3} \ne \frac{1^2}{1^2}\) as well.

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You have \(\displaystyle \frac{3}{x^2+3^2} = \frac{3}{3}\cdot\frac{1}{\frac{x^2}{3}+\frac{3^2}{3}}\)

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

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

 

[I like Serena]

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

Huh? No.

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

 

[Petrus]

Huh? No.

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

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

 

[I like Serena]

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

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 \(\displaystyle u=\frac{x}{3}\).
This means that \(\displaystyle du=\frac{dx}{3}\).

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 \(\displaystyle u=\frac{x}{3}\).
This means that \(\displaystyle du=\frac{dx}{3}\).

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 \(\displaystyle [\arctan{u}]_{\frac{4n-1}{6}}^{\frac{12n+1}{6}}\) we get and then calculate limit we get \(\displaystyle \arctan(infinity)-arctan(infinty)\) the answer shall be \(\displaystyle \arctan(6)-\arctan(2)\) I am doing something wrong?

 

[I like Serena]

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

 

[Opalg]

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

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