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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,

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

From this answer here,

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

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
tryst with freedom
  • 11,538

1 Answers1

2

A quick read of the proof did not show any error, although I would need to take a good look into it. The result is indeed correct, so I hope there is no major error.

For the second question, there is indeed a more efficient proof of this identity.

\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 $\sum_{i=0}^nS(n,i)(k)_i=k^j$.

As for simplifications, nothing really jumps out beyond this. Given that there are alternate representations of this summation using Euler numbers and Bernoulli numbers (which are special numbers with no known closed forms), it is highly suggestive that one cannot get this into any good closed form.

I will keep working on this problem and add anything if I find more.

Yuzuriha Inori
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