# Faulhaber formula from geometric series and operators?

As shown in this post, $$\sum_{k=1}^n x^k = x \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1}$$

For RHS, notice $$x= \left(1+( x-1) \right)$$ and using this we get,

$$\sum_{k=1}^n x^k = \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1} + \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k} \tag{1}$$

For first term,

$$\sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1} \to \binom{n}{1} +\sum_{k=2}^{n} \binom{n}{k} (x-1)^{k-1} \tag{2}$$

Sub, $$k-1 \to j$$

$$\sum_{k=2}^{n} \binom{n}{k} (x-1)^{k-1} \to + \sum_{j=1}^{n-1} \binom{n}{j+1} (x-1)^j \to + \sum_{k=1}^{n-1} \binom{n}{k+1} (x-1)^k \tag{3}$$

Using (1), (2), and (3)

$$\sum_{k=1}^n x^k = \binom{n}{1} + \sum_{k=1}^{n-1} \binom{n}{k+1} (x-1)^k + \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k}$$

Or,

$$\sum_{k=1}^n x^k= \binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^{n}$$ =

Now apply the $$P^j$$ to both sides (4) where $$P$$ is an operator defined as $$x \frac{d}{dx}$$ and evaluate at x=1, see this post for more details. For LHS,

$$\sum_{k=1}^n x^k \xrightarrow[]{P^j , x=1} \sum_{k=1}^n k^j$$

$$P^j =\sum_{i=1}^j S(j,i) D_{1}^i$$

Where $$D_1^i = \frac{d^i}{dx^i}|_{x=1}$$ and S(n,k) is stirling number of second kind

Writing (4) out explicitly,

$$\sum_{k=1}^n k^j = \sum_{i=1}^j S(j,i) D_{1}^i \left[ \binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^n \right]$$

Now, consider

$$D_{1}^i \left[\binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^n \right]$$

We can easily evaluate this by considering taylor series of the inside term, call:

$$f= \binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^n$$

Then, the taylor polynomial of $$f$$ around $$x=1$$ is given as:

$$f = \sum_{k=0}^{n+1} \frac{d^k f}{dx^k}|_1 \frac{(x-1)^k}{k!}$$

By comparing coefficients we can easily evaluate the derivative,

$$D_{1}^i \left[\binom{n}{1}+ \sum_{k=1}^{n-1} \binom{n+1}{k+1} (x-1)^k + (x-1)^n \right] = \begin{cases} \binom{n}{0} , i=0 \\ i! \binom{n+1}{i+1} , i>0 \end{cases}$$

For $$i \in \mathbb{N}$$, hence:

$$\sum_{k=1}^n k^j = \sum_{i=1}^j S(j,i) i! \binom{n+1}{i+1}$$

With all of that in mind,

1. Is my proof right?
2. What ways can I make it better?
3. Are there any more simplification applicable?
Note: I am evaluating the quantity operator on by $$P^j$$ at x=1
• I haven't followed the entire proof, but the second line has an error which continues throughout. The upper limit on the LHS should be n+1 not n. Nov 22 '20 at 23:22
• fixed it I think Nov 23 '20 at 7:11

\begin{align*} \sum_{k=0}^nk^j&=\sum_{k=0}^n\sum_{i=0}^j S(j,i)(k)_i\\&=\sum_{i=0}^j i!S(j,i)\sum_{k=0}^n\frac{(k)_i}{i!}\\&=\sum_{i=0}^j i!S(j,i)\sum_{k=0}^n\binom{k}{i}\\&=\sum_{i=0}^j i!S(j,i)\binom{n+1}{i+1} \end{align*}
where $$(k)_i$$ is the falling factorial and in the first equality, I have used the fact that $$\sum_{i=0}^nS(n,i)(k)_i=k^j$$.