Summation of roots In this post, Calculate summation of square roots we are shown how to sum square roots. My question is, can we get similarly simple expressions if instead of square roots we choose some other exponent. More specifically, is there a simple approximating expression for
$\sum_{i=1}^N i^{2/3}$
 A: Well writing $\;\displaystyle\sum_{k=1}^N k^a=H_N^{\left(-a\right)}\;$ is just a rewriting using generalized harmonic numbers.
Writing this as a partial zeta sum or truncated zeta function $\zeta_N(-a)$ is exactly the same thing!
You get too $\;\displaystyle\sum_{k=1}^N k^a =\zeta(-a) - \zeta(-a, N+1)\;$ using the Hurwitz zeta function.
For $a$ integer you get Faulhaber's formula and for real or even complex values $\,a\neq -1\,$ excellent approximations are provided by the Euler-Maclaurin formula (with $s=-a$) :
$$\sum_{k=1}^N k^a \sim\zeta(-a)+\frac {N^{a+1}}{a+1}+\frac {N^a}2-\sum_{i\ge 1}\frac{B_{2i}}{(2i)!}\frac{(s+2i-2)!\;N^{a-2i+1}}{(s-1)!}$$
(note that this is an asymptotic expansion and that the sum at the right has to remain finite with $B_j$ a Bernoulli numbers)
A: For (increasing) concave functions $f$ with $f(x)\to\infty$ as $x\to\infty,$ for example $f(x) = x^{2/3},$ integrating i.e. finding the area under the curve between $1$ and $a,$ $\displaystyle\int_1^a f(x)\ dx,$ can give you a relatively good approximation to the sum $\displaystyle\sum_{n=1}^{a} f(n)$, but not a good absolute difference. If you want to be more precise then you have to do a bit more work.
For (increasing) concave functions $f$ with $f(x)\to c\in\mathbb{R}$ as $x\to\infty,$ for example $f(x) = 1-1/x,$ integrating between $1$ and $a,$ $\displaystyle\int_1^a f(x)\ dx,$ not only gives you a good relative approximation but in fact can give you a good absolute approximation to $\displaystyle\sum_{n=1}^{a} f(n)$: in the case of $f(x)=1-1/x$ the area under the curve (i.e. the integral) differs from the sum $\sum_n 1/n$ by a maximum of some constant which is probably less than $2$.
Seeing these things is not so difficult once you've drawn diagrams to see what's going on.
