Geometric Sum (Power Series) Calculation

1. Mar 28, 2008

rwinston

Hi

In trying to calculate the following sum:

$$\sum_{i=1}^n{i^2}$$

I found the following expansions:

$$\sum_{i=1}^p \sum_{j=0}^{i-1} (-1)^j(i-j)^p {n+p-i+1\choose n-i} {p+1\choose j}$$

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

2. Mar 28, 2008

Pere Callahan

Yes, it is infinity.

And what you have is not a geometric series.

Last edited: Mar 28, 2008
3. Mar 28, 2008

tiny-tim

Hi rwinston!

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

4. Mar 28, 2008

flebbyman

5. Mar 28, 2008

rwinston

Hi tim

So I think the expansion of that sum is

$$\sum_{i=1}^n{i^3+3i^2+3i+1} - \sum_{i=1}^n{i^3}$$
$$=\sum_{i=1}^{n}{3i^2+3i+1}$$
$$=3\sum_{i=1}^n{i^2}+3\sum_{i=1}^n{i}+n$$

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

Last edited: Mar 28, 2008
6. Mar 28, 2008

tiny-tim

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!