I would like to show that for $s \in \mathbb{R}$ and a nonnegative integer $k$ $$ \triangle^k ((1+n)^s)) \lesssim (1+|n|)^{s-k} $$

where $\triangle$ is the discrete derivative, i.e. $\triangle^1 ((1+n)^s)) = (2+n)^s - (1+n)^s$.

This is easy when $s \in \mathbb{Z}$, and in the continuous analogue because $$ \partial_x^k (1+x)^s = (s)(s-1)\cdots (s-k) (1+x)^{s-k} $$

I think that you can use the generalized binomial theorem to prove this, but I was wondering if there was anything more straightforward, e.g. some kind of convexity argument to use the continuous case.

Note: I wasn't sure about the tags, feel free to re-tag as appropriate.

I would like to show that for $s \in \mathbb{R}$ and a nonnegative integer $k$ $$ \triangle^k ((1+n)^s) \lesssim (1+|n|)^{s-k} $$

where $\triangle$ is the discrete derivative, i.e. $\triangle^1 ((1+n)^s) = (2+n)^s - (1+n)^s$.

This is easy when $s \in \mathbb{Z}$, and in the continuous analogue because $$ \partial_x^k (1+x)^s = (s)(s-1)\cdots (s-k+1) (1+x)^{s-k} $$

I think that you can use the generalized binomial theorem to prove this, but I was wondering if there was anything more straightforward, e.g. some kind of convexity argument to use the continuous case.

Note: I wasn't sure about the tags, feel free to re-tag as appropriate.