Infinite Series and Limits

**metallicsatan** · November 10th 2010, 01:04 AM

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?

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

**tonio** · November 10th 2010, 01:32 AM

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

**Also sprach Zarathustra** · November 10th 2010, 02:18 AM

.

**metallicsatan** · November 10th 2010, 06:04 AM

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

**tonio** · November 10th 2010, 07:43 AM

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

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

Tonio

**metallicsatan** · November 10th 2010, 08:57 AM

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?

**tonio** · November 10th 2010, 09:11 AM

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

**metallicsatan** · November 10th 2010, 09:55 AM

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.

**HallsofIvy** · November 10th 2010, 10:07 AM

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

**metallicsatan** · November 10th 2010, 10:13 AM

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

**Also sprach Zarathustra** · November 10th 2010, 11:35 AM

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

**metallicsatan** · November 10th 2010, 11:58 AM

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?

**metallicsatan** · November 10th 2010, 12:00 PM

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?

**tonio** · November 10th 2010, 06:28 PM

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.

.

**ineedyourhelp** · November 10th 2010, 10:56 PM

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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