# Infinite Series and Limits

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• Nov 10th 2010, 01:04 AM
metallicsatan
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? :)

$\displaystyle \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{n^2}{n^2+k^2}$
• Nov 10th 2010, 01:32 AM
tonio
Quote:

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? :)

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

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

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

And in what interval, of course?

Tonio
• Nov 10th 2010, 02:18 AM
Also sprach Zarathustra
.
• Nov 10th 2010, 06:04 AM
metallicsatan
Quote:

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

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

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?
$\displaystyle \displaystyle \lim_{k \to \infty}\sum_{n=1}^k \frac{1}{1+\left(\frac{n}{k}\right)^2} = \displaystyle \sum_{n=1}^\infty 1 = \infty$
• Nov 10th 2010, 07:43 AM
tonio
Quote:

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

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

Tonio
• Nov 10th 2010, 08:57 AM
metallicsatan
I can only think of $\displaystyle \frac{1}{1+x^2}$ but the problem is, the summation does not have a partition, if that is the case. Any more hints?
• Nov 10th 2010, 09:11 AM
tonio
Quote:

Originally Posted by metallicsatan
I can only think of $\displaystyle \frac{1}{1+x^2}$ 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
• Nov 10th 2010, 09:55 AM
metallicsatan
so is $\displaystyle \Delta x=1$? and $\displaystyle I=[0,k]$. If that is the case, the limit is equal to $\displaystyle \int_{0}^{\infty}\frac{1}{1+x^2}\, dx=\frac{\pi}{2}$? Right? I'm not sure. (Thinking)
• Nov 10th 2010, 10:07 AM
HallsofIvy
Quote:

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

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

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

Quote:

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

And in what interval, of course?

Tonio
• Nov 10th 2010, 10:13 AM
metallicsatan
Now I have totally no idea what to do. Any more hints?
• Nov 10th 2010, 11:35 AM
Also sprach Zarathustra
In my poor opinion this problem can be solved without using Riemann sums... ( Thinking in a new direction)
• Nov 10th 2010, 11:58 AM
metallicsatan
Quote:

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?
• Nov 10th 2010, 12:00 PM
metallicsatan
Quote:

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

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

using this Riemann integral, what should be the partition and the boundaries?
• Nov 10th 2010, 06:28 PM
tonio
Quote:

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

Oops! I totally missed that typo...(Giggle)

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 $\displaystyle \frac{1}{1+ x^2}$.

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

.
• Nov 10th 2010, 10:56 PM
ineedyourhelp
Quote:

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

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

Hi, if we change it to riemann sum, should there be kX1/k introduced as well? so we will end up with $\displaystyle k\int \frac{x^2}{1+ x^2}$dxwhere do the k go from here?
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