General formula for the sum of a finite power series

[mjpam]

I was wondering if there was a general way to find the sum of a finite power series:

$$\sum_{n=1}^{N}{n^{m}}$$

where $m$ is a fixed integer.

Now, there is some math folklore that a seven- (or ten-)year-old Gauss solved the $m=1,\;N=100$ case by realizing that by reversing the series and summing the respective terms in the each series, he got 101 added together 100 time so that all he had to do the get the sum of the original series was to divide by 2.

Symbolically and more generally the procedure is:

$$\sum_{n=1}^{N}n=\underset{\textup{N terms}}{\underbrace{1+2+\cdots+(N-1)+N}}$$
$$\sum_{n=1}^{N}n=\underset{\textup{N terms}}{\underbrace{N+(N-1)+\cdots+2+1}}$$
$$2\sum_{n=1}^{N}n=(N+1)+((N-1)+2)+\cdots+(2+(N-1))+(1+N)$$
$$2\sum_{n=1}^{N}n=(N+1)+(N+1)+\cdots+(N+1)+(N+1)$$
$$2\sum_{n=1}^{N}n=N(N+1)$$
$$\sum_{n=1}^{N}n=\frac{N(N+1)}{2}$$

Now this reduce to a relatively simple formula because each of the respective terms in the forward and backward series sums to the same value. This is however not the case with the general power series:

$$\sum_{n=1}^{N}n^{m}=\underset{\textup{N terms}}{\underbrace{1^{m}+2^{m}+\cdots+(N-1)^{m}+N^{m}}}$$
$$\sum_{n=1}^{N}n^{m}=\underset{\textup{N terms}}{\underbrace{N^{m}+(N-1)^{m}+\cdots+2^{m}+1^{m}}}$$
$$2\sum_{n=1}^{N}n^{m}=(N^{m}+1^{m})+((N-1)^{m}+2^{m})+\cdots+(2^{m}+(N-1)^{m})+(1^{m}+N^{m})}$$

Is the a way to express the sum of a general finite power series in terms of the exponent and the number of terms in the series? DO you have to use the binomial expansion?

 

[disregardthat]

Yes, there is, but it's complicated. This is how you find it recursively:

Let $$S(n,m) = \sum^n_{k=1} k^m$$.

Then
$$(n+1)^m = S(n+1,m)-S(n,m) = \sum^{n+1}_{k=1} k^m - \sum^n_{k=1} k^m = 1+\sum^n_{k=1} (k+1)^m-k^m = 1+\sum^n_{k=1} \sum^{m-1}_{r=0} {m \choose r} k^r = 1+ \sum^{m-1}_{r=0} {m \choose r} \sum^n_{k=1} k^r = 1+ \sum^{m-1}_{r=0} {m \choose r} S(n,r)$$

so $$S(n,m-1) = \frac{(n+1)^m-1-\sum^{m-2}_{r=0}{m \choose r} S(n,r)}{m}$$

i.e. $$S(n,m) = \frac{(n+1)^{m+1}-1-\sum^{m-1}_{r=0}{m+1 \choose r} S(n,r)}{m+1}$$


For $$m \geq 1$$.

The formula itself is not easy to derive, but this is it: http://en.wikipedia.org/wiki/Faulhaber's_formula

So you have a closed form formula for the sum.