# Thread: Infinite Series and Limits

1. ## Infinite Series and Limits

This was a problem in my test this morning. I was not able to answer it. But I thought it was pretty interesting, and would like to know the answer. Could you guys help me out?

$\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}$

2. Originally Posted by metallicsatan
This was a problem in my test this morning. I was not able to answer it. But I thought it was pretty interesting, and would like to know the answer. Could you guys help me out?

$\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}$

$\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}=\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2}$ .

To what familiar and very easy to integrate function's Riemann sum does the last sum above ressemble?

And in what interval, of course?

Tonio

3. .

4. Originally Posted by tonio
$\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}=\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2}$ .

To what familiar and very easy to integrate function's Riemann sum does the last sum above ressemble?

And in what interval, of course?

Tonio
That is just the given in the problem. I don't know if it's supposed to be a Riemman sum of some function over an integral.

Also, is the following statement correct?
$\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2} =
\displaystyle \sum_{n=1}^\infty 1 = \infty$

5. Originally Posted by metallicsatan
That is just the given in the problem. I don't know if it's supposed to be a Riemman sum of some function over an integral.

Also, is the following statement correct?
$\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2} =
\displaystyle \sum_{n=1}^\infty 1 = \infty$

No, it isn't correct. And you have to figure out the function, the interval and its partition. Think it hard.

Tonio

6. I can only think of $\frac{1}{1+x^2}$ but the problem is, the summation does not have a partition, if that is the case. Any more hints?

7. Originally Posted by metallicsatan
I can only think of $\frac{1}{1+x^2}$ but the problem is, the summation does not have a partition, if that is the case. Any more hints?

That's the function...now what's the interval and its partition? Check carefully the definition of Riemann sum...

Tonio

8. so is $\Delta x=1$? and $I=[0,k]$. If that is the case, the limit is equal to $\int_{0}^{\infty}\frac{1}{1+x^2}\, dx=\frac{\pi}{2}$? Right? I'm not sure.

9. Originally Posted by tonio
$\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}=\displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2}$ .
This is incorrect. Dividing both numerator and denominator of $\frac{n^2}{n^2+ k^2}$ by $n^2$ gives $\frac{1}{1+ \left(\frac{k}{n}\right)^2}$. And, now, unfortunately, that is NOT the same as [/tex]\frac{1}{1+ x^2}[/tex].

Perhaps dividing both numerator and denominator by $k^2$ giving $\frac{\left(\frac{n}{k}\right)^2}{1+ \left(\frac{n}{k}\right)^2}$ so we can think of it as a Riemann sum for $\int \frac{x^2}{1+ x^2} dx= \int x-\frac{x}{x^2+ 1}$ will work.

To what familiar and very easy to integrate function's Riemann sum does the last sum above ressemble?

And in what interval, of course?

Tonio

10. Now I have totally no idea what to do. Any more hints?

11. In my poor opinion this problem can be solved without using Riemann sums... ( Thinking in a new direction)

12. Originally Posted by Also sprach Zarathustra
In my poor opinion this problem can be solved without using Riemann sums... ( Thinking in a new direction)
how?

13. Originally Posted by HallsofIvy
This is incorrect. Dividing both numerator and denominator of $\frac{n^2}{n^2+ k^2}$ by $n^2$ gives $\frac{1}{1+ \left(\frac{k}{n}\right)^2}$. And, now, unfortunately, that is NOT the same as [/tex]\frac{1}{1+ x^2}[/tex].

Perhaps dividing both numerator and denominator by $k^2$ giving $\frac{\left(\frac{n}{k}\right)^2}{1+ \left(\frac{n}{k}\right)^2}$ so we can think of it as a Riemann sum for $\int \frac{x^2}{1+ x^2} dx= \int x-\frac{x}{x^2+ 1}$ will work.
using this Riemann integral, what should be the partition and the boundaries?

14. Originally Posted by HallsofIvy
This is incorrect. Dividing both numerator and denominator of $\frac{n^2}{n^2+ k^2}$ by $n^2$ gives $\frac{1}{1+ \left(\frac{k}{n}\right)^2}$.

Oops! I totally missed that typo...

The problem is that k is the one going to infinity...I did a mess with the indexes.

Oh, well...

Tonio

And, now, unfortunately, that is NOT the same as $\frac{1}{1+ x^2}$.

Perhaps dividing both numerator and denominator by $k^2$ giving $\frac{\left(\frac{n}{k}\right)^2}{1+ \left(\frac{n}{k}\right)^2}$ so we can think of it as a Riemann sum for $\int \frac{x^2}{1+ x^2} dx= \int x-\frac{x}{x^2+ 1}$ will work.

.

15. Originally Posted by HallsofIvy
This is incorrect. Dividing both numerator and denominator of $\frac{n^2}{n^2+ k^2}$ by $n^2$ gives $\frac{1}{1+ \left(\frac{k}{n}\right)^2}$. And, now, unfortunately, that is NOT the same as [/tex]\frac{1}{1+ x^2}[/tex].

Perhaps dividing both numerator and denominator by $k^2$ giving $\frac{\left(\frac{n}{k}\right)^2}{1+ \left(\frac{n}{k}\right)^2}$ so we can think of it as a Riemann sum for $\int \frac{x^2}{1+ x^2} dx= \int x-\frac{x}{x^2+ 1}$ will work.
Hi, if we change it to riemann sum, should there be kX1/k introduced as well? so we will end up with $k\int \frac{x^2}{1+ x^2}$dxwhere do the k go from here?

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