TITLE: Sum of all consecutive natural root differences on a given power
QUESTION [10 upvotes]: I accidentally observed that $\sqrt{n} - \sqrt{n-1}$ tends to $0$ for higher values of $n$, so I've decided to try to sum it all, but that sum diverged. So I've tried to make it converge by giving it a power $m$.

$$S_m=\sum_{n=1}^\infty (\sqrt{n} - \sqrt{n-1})^m$$
How would one calculate the values of this sum for a choosen $m\in\mathbb R$?
Not just estimate but write them in a non-decimal form if possible, preferably using a better converging formula.

It converges if $m>2$. 
The values seem to tend to numbers with non-repeating decimals.

Thanks to achille hui and his partial answer here, it looks like $S_m$ for odd values of $m$ is a linear combination
of riemann zeta function at negative half-integer values:  

\begin{align} S_3 &\stackrel{}{=} -6\zeta(-\frac12) \approx 1.247317349864128...\\ S_5 &\stackrel{}{=} -40\zeta(-\frac32) \approx 1.019408075593322...\\ S_7 &\stackrel{}{=} -224\zeta(-\frac52) - 14\zeta(-\frac12) \approx 1.00261510344449... \end{align}

If we decide to replace the constant $m$ with $n\times{k}$ where $k$ is a new constant, then we can talk about $S_k$, Which converges if $k>0$;

$$S_k=\sum_{n=1}^\infty (\sqrt{n} - \sqrt{n-1})^{nk}$$
I wonder if these values could also be expressed in a similar way as $S_m$.

Values still tend to numbers with seemingly non-repeating decimals according to the Wolfram Alpha:
$$ S_1 \approx 1.20967597004937847717395464774494290
$$

Also notice that the functions of these sums are similar to the zeta function; $\color{blue}{S_m} \sim \color{red}{\zeta}$

But I think it's only due the fact that they all approach $1$?

REPLY [3 votes]: It does converge for $m > 2$, since $\sqrt{n} - \sqrt{n-1} \sim 1/(2\sqrt{n})$ and $\sum_n n^{-m/2}$ converges for $m > 2$.