1. ## Integration Limit

Hi. I was thinking a bit and I tried to use the definition of differentiation to see how the rules were created like the standard "multiply by the power and minus one from the power" for functions in the form of x^a. That was easy enough.

I then tried to do the same thing with integration using x^a again which should lead me to x^(a+1) / (a+1). I got stuck at a point when testing it with x^2.

Limit n-> Infinity of ( (1 + 2^2 + 3^2 + 4^2 + ... n^2) / n^3 )

How do I solve this limit algebraically?

2. $\displaystyle 1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

Hence, your limit clearly: $\displaystyle \frac{1}{3}$

3. What you have is:

$\displaystyle \displaystyle\frac{1}{n^3}\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6n^3}$

I have used this Square pyramidal number - Wikipedia, the free encyclopedia. I'm sure you can find the limit of the above to be 1/3 which would be along what you are doing. I think you are using the Riemann sum definition? Lets try to find the integral from 0 to x. We pick the right-most points of the intervals, which are of length $\displaystyle x/n$ and we get

$\displaystyle \displaystyle \sum_{k=1}^n f\left(k\frac{x}{n}\right)\frac{x}{n} = \sum_{k=1}^n \left(\frac{k x}{n}\right)^a \frac{x}{n} = \frac{x^{a+1}}{n^{a+1}}\sum_{k=1}^n k^a$

It seems like we are stuck but there is a formula you could use to take it from there. With this formula you can find the derivative exactly.

EDIT: oops, too late with that first part. Oh, well, at least you have the more general setting

4. $\displaystyle 1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

Getting the sum in that form is exactly what i needed :P I have difficulty seeing how to translate a series of sums into a form like that.

5. Originally Posted by Corpsecreate
$\displaystyle 1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$

Getting the sum in that form is exactly what i needed :P I have difficulty seeing how to translate a series of sums into a form like that.
Oh, those are difficult! In my school we had to learn the first few powers (1,2 and 3 I think). In general, there is no closed form. however, there is a formula that will help you calculate those sums. It's called the Faulhaber's formula and you can learn more about it here -> Faulhaber's formula - Wikipedia, the free encyclopedia. If you use this formula and take the leading term from the sum, you can then use it on the above integral and take the limit without calculating all those tricky bernoulli numbers that it involves. If you need to see the sum with its leading term taken out of the sum, here is what it looks like: Summation - Wikipedia, the free encyclopedia

6. I love how seemingly impossible sums can be expressed in something so neat and compact. Thanks for the help and the info.

7. I will show you how to get the formula for the first $\displaystyle n$ terms in:

$\displaystyle 1^5+2^5+3^5+...+n^5$

First, we notice (prove it) that the above sum equals to some polynomial with degree $\displaystyle 6$.

In other words:

(1) $\displaystyle 1^5+2^5+3^5+...+n^5=An^6+Bn^5+Cn^4+Dn^3+En^2+Fn+G$

Our objective is to find the coefficients $\displaystyle A,B,C,D,E,F,G$.

Solution:

Let $\displaystyle n=1$, then (1) becomes:

$\displaystyle 1=A+B+C+D+E+F+G$

It's our first linear equation in 7 variables, so we need more 6 equations:

Let now $\displaystyle n=2$, then (1) becomes:

$\displaystyle 1^5+2^5=2^6A+2^5B+2^4C+2^3D+2^2E+2^1F+G$

or:

$\displaystyle 33=64A+32B+16C+8D+4E+2F+G$

The other five equations you will get in similar way...

Now, by solving a linear system of equations in 7 variables you will get: $\displaystyle A,B,C,D,E,F,G$.(It's the hard part, but it's standard...)