Geometric Sum (Power Series) Calculation

[rwinston]

Hi

In trying to calculate the following sum:

<br /> \sum_{i=1}^n{i^2}<br />

I found the following expansions:

<br /> \sum_{i=1}^p \sum_{j=0}^{i-1} (-1)^j(i-j)^p {n+p-i+1\choose n-i} {p+1\choose j}<br />

My question is: is there an easier or more intuitive way to compute the limit of the sum above?

 

[Pere Callahan]

Yes, it is infinity.

And what you have is not a geometric series.

 

[tiny-tim]

Hi rwinston!

Hint: what is \sum_{i=1}^n{\left((i+1)^3\,-\,i^3}\right) ?

 

[flebbyman]

Hey, here you go: http://en.wikipedia.org/wiki/Summation#Identities

 

[rwinston]

Hi tim

So I think the expansion of that sum is

<br /> \sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}<br />
<br /> =\sum_{i=1}^{n}{3i^2+3i+1}<br />
<br /> =3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n<br />

Im not sure what the connection is yet tho...

 

[tiny-tim]

Hi tim

So I think the expansion of that sum is

<br /> \sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}<br />
<br /> =\sum_{i=1}^{n}{3i^2+3i+1}<br />
<br /> =3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n<br />

Im not sure what the connection is yet tho...

ah… but you know what 3\sum_{i=1}^n{i}\,+\,n is;

And you also know what \sum_{i=1}^n{\left((i+1)^3\,-\,i^3}\right) is … don't you … ?

So \sum_{i=1}^n{i^2} is one-third the difference!