# How to obtain a closed form for summation over polynomial ($\sum_{x=1}^n x^m$)?

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• Explicit formula for Bernoulli numbers by using only the recurrence relation

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#### Solutions Collecting From Web of "How to obtain a closed form for summation over polynomial ($\sum_{x=1}^n x^m$)?"

Here is an alternate formulation that is slightly different from
Faulhaber’s formula.

Suppose we are trying to evaluate
$$S_m(n) = \sum_{k=1}^n k^m$$
where $m$ is a positive integer.

Because $m$ is positive we may extend this to include zero.

As I have suggested at this MSE link,
put $$k^m = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \exp(kw) \; dw \\ = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \sum_{q=0}^k {k\choose q} (\exp(w)-1)^q \; dw \\ = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \sum_{q=0}^k \frac{k!}{(k-q)!} \frac{(\exp(w)-1)^q}{q!} \; dw \\ = \sum_{q=0}^k \frac{k!}{(k-q)!} \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \frac{(\exp(w)-1)^q}{q!} \; dw \\ = \sum_{q=0}^k \frac{k!}{(k-q)!} {m\brace q}.$$

This yields for the sum
$$\sum_{k=0}^n \sum_{q=0}^k \frac{k!}{(k-q)!} {m\brace q} = \sum_{q=0}^n {m\brace q} \sum_{k=q}^n \frac{k!}{(k-q)!} \\ = \sum_{q=0}^n q! {m\brace q} \sum_{k=q}^n {k\choose q}.$$

Now observe that when $n>m$ the Stirling number is zero for $q>m$, so we may
may set the upper limit on the outer sum to $m$ without loosing any terms. On the other hand when $n<m$ the inner sum is zero when $q>n$ so we may again set the upper limit to $m.$

This yields
$$\sum_{q=0}^m q! {m\brace q} \sum_{k=q}^n {k\choose q}.$$

To evaluate the inner sum put
$${k\choose q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^k}{z^{q+1}} \; dz.$$

This yields for the sum
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \sum_{k=q}^n (1+z)^k\; dz = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \frac{(1+z)^{n+1}-(1+z)^q}{1+z-1}\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+2}} ((1+z)^{n+1}-(1+z)^q)\; dz.$$

With $n\ge q$ this evaluates to
$${n+1\choose q+1}$$
by inspection.

Returning to the main thread we have established that
$$S_m(n) = \sum_{q=0}^m q! {m\brace q} {n+1\choose q+1}.$$

The Stirling number starts at $q=m$ and thus we have a polynomial in
$n$ of degree $m+1.$

Recall the OGF of the signed Stirling numbers of the first kind which
is
$$\sum_{k=1}^n (-1)^{n-k} {n\brack k} z^k = n! \times {z\choose n}.$$

It follows that
$$[n^p] {n+1\choose q+1} = [n^p] \frac{n+1}{q+1} {n\choose q} = \frac{1}{q+1} \left([n^{p-1}] {n\choose q} + [n^{p}] {n\choose q} \right) \\ = \frac{1}{(q+1)!} \left((-1)^{q-p} {q\brack p} + (-1)^{q-p-1} {q\brack p-1} \right) \\ = \frac{(-1)^{q-p}}{(q+1)!} \left({q\brack p} – {q\brack p-1} \right).$$

We thus have
$$[n^p] S_m(n) = \sum_{q=0}^m q! {m\brace q} \frac{(-1)^{q-p}}{(q+1)!} \left({q\brack p} – {q\brack p-1} \right) \\ = \sum_{q=0}^m {m\brace q} \frac{(-1)^{q-p}}{q+1} \left({q\brack p} – {q\brack p-1} \right).$$

This finally gives the desired polynomial in $n$ for $S_n(m):$
$$S_m(n) = \sum_{p=1}^{m+1} n^p \sum_{q=0}^m {m\brace q} \frac{(-1)^{q-p}}{q+1} \left({q\brack p} – {q\brack p-1} \right).$$

Hint: Write an equation with $x^{m+1} – (x-1)^{m+1}$ on the LHS and its expansion using the binomial theorem on the RHS. Now apply the summation operator to both sides, and the LHS telescopes. Rearrange to get $\sum x^m$ as a sum of $\sum x^k$ for each $k < m$.

This proves that $\sum x^m$ is a polynomial. You can then use polynomial interpolation to construct the polynomial.