$\Delta^kn^\alpha$ converges monotonically to zero when $\alphaI am trying to prove that the $k$-th finite difference of the series $n^\alpha$ converges to zero monotonically as $n\to\infty$ when $\alpha<k$. The differential analogue is
$$\frac{d^k}{dx^k}x^\alpha=x^{\alpha-k}\alpha(\alpha-1)\cdots(\alpha-k+1)$$
which goes to zero as $x\to\infty$ when $\alpha<k$ because $x^y\to0$ for $y<0$, and of course monotonicity is trivial.
The definition of the finite difference series is $\Delta^0n^\alpha=n^\alpha$, $\Delta^kn^\alpha=\Delta^{k-1}n^\alpha-\Delta^{k-1}(n+1)^\alpha$, or in closed form as
$$\Delta^kn^\alpha=\sum_{m=0}^k(-1)^m{k\choose m}(n+m)^\alpha,$$
and it is a rough analogue of the derivative for series (note that this definition has an extra minus sign by comparison to the usual definition, although it doesn't affect the theorem), so I expect that the fact that this is true for derivatives on the continuous function $x^\alpha$ should carry over to the series. Indeed, for $\alpha$ a nonnegative integer, the theorem follows from the observation that when $p(n)$ is a degree $q$ polynomial, $\Delta p(n)$ is at most degree $q-1$ since $a_q(n+1)^q-a_qn^q=a_q\sum_{m=0}^{q-1}{q\choose m}n^m$ which has no degree $q$ components, so $\Delta^{q+1}p(n)=0$ and more generally $\Delta^kp(n)=0$ for $k>q$. Can this proof be generalized to noninteger real $\alpha$?
 A: Note that
$$
(n+1)^{\alpha}-n^{\alpha} = \alpha\int_{n}^{n+1}x^{\alpha-1}dx
$$
and
$$
((n+2)^{\alpha}-(n+1)^{\alpha})-((n+1)^{\alpha}-n^{\alpha}) = \alpha\int_{n}^{n+1}(x+1)^{\alpha-1}dx-\alpha\int_{n}^{n+1}x^{\alpha-1}dx \\
\alpha\int_{n}^{n+1}((x+1)^{\alpha-1}-x^{\alpha-1})dx = \alpha(\alpha-1)\int_{n}^{n+1}\int_{x}^{x+1}y^{\alpha-2}dy
$$
and by induction, for instance, in general
$$
\Delta^{k}n^{\alpha} = \alpha(\alpha-1)\cdots(\alpha-k+1)\int_{n}^{n+1}\int_{x_1}^{x_1+1}\int_{x_2}^{x_2+1}\cdots\int_{x_{k-1}}^{x_{k-1}+1}x_{k}^{\alpha-k}dx_{k}dx_{k-1}\ldots dx_{2}dx_{1}.
$$
Now, just define
$$
f_{i}(x) = \int_{x}^{x+1}\int_{x_1}^{x_1+1}\int_{x_2}^{x_2+1}\cdots\int_{x_{j-1}}^{x_{j-1}+1}x_{j}^{\alpha-k}dx_{j}dx_{j-1}\ldots dx_{2}dx_{1}
$$
and note that
$$
f_{i+1}(x) = \int_{x}^{x+1}f_{i}(t)dt \Rightarrow f'_{i+1}(x) = f_{i}(x+1)-f_{i}(x)
$$
and since $f_{0}(x) = x^{\alpha-k}$ is decreasing when $\alpha < k$, by induction all $f_{i}(x)$ are. In particular sequence of $k$:th finite differences is monotonic. But now
$$
f_{i+1}(x) = \int_{x}^{x+1}f_{i}(t)dt \leq ((x+1)-x)f_{i}(x) = f_{i}(x)
$$
so $f_{k}(x) \leq f_{0}(x) = x^{\alpha-k} \to 0$ as $x \to \infty$, so also $k$:th finite differences converge monotonically to zero.
