What is $\sum\limits_{i=1}^n \sqrt i\ $? What is $\sum\limits_{i=1}^n\sqrt i\ $? 
Also I noticed that $\sum\limits_{i=1}^ni^k=P(n)$
where $k$ is a natural number and $P$ is a polynomial of degree $k+1$. Does that also hold for any real positive number? How could one prove it?
 A: Using the Euler-Maclaurin Sum Formula, we get the asymptotic expansion for $\displaystyle\sum_{i=1}^n\sqrt{i}$ :
$$
\frac23n^{3/2}+\frac12n^{1/2}+\zeta\left(-\frac12\right)+\frac1{24}n^{-1/2}-\frac1{1920}n^{-5/2}+\frac1{9216}n^{-9/2}+O\left(n^{-13/2}\right)
$$
where $\zeta\left(-\frac12\right)=-\frac{1}{4\pi}\zeta\left(\frac32\right)\doteq-0.20788622497735456602$.

As described in this answer, for $\mathrm{Re}(z)\gt-1$,
$$
\zeta(z)=\lim_{n\to\infty}\left(\sum_{k=1}^nk^{-z}\;-\;\left(\frac{1}{1-z}n^{1-z}+\frac12n^{-z}\right)\right)
$$
The formula above agrees for $\mathrm{Re}(z)\gt1$ and the limit is analytic.
Letting $z=-\frac12$, yields that the constant in the Euler-Maclaurin Sum Formula is $\zeta\left(-\frac12\right)$.
A: $$
\sum_{i=1}^n\sqrt{i}=f(n)
$$
$$
\sum_{i=1}^{n-1}\sqrt{i}=f(n-1)
$$
$$
f(n)-f(n-1)=\sqrt{n}
 \tag 1$$
We know Taylor expansion
$$
f(x+h)=f(x)+hf'(x)+\frac{h^2 f''(x)}{2!}+\frac{h^3f'''(x)}{3!}+....
$$
Thus
$$
f(n-1)=f(n)-f'(n)+\frac{f''(n)}{2!}-\frac{f'''(n)}{3!}+....
$$
$$
f(n)-f(n)+f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-....=\sqrt{n}
$$
$$
f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-...=\sqrt{n}
$$
$$
f(n)-\frac{f'(n)}{2!}+\frac{f''(n)}{3!}-...=\int \sqrt{n} dn
$$
$$
f(n)-\frac{f'(n)}{2!}+\frac{f''(n)}{3!}-\frac{f'''(n)}{4!}...=\frac{2}{3}n^\frac{3}{2} +c
$$
$$
\frac{1}{2} ( f'(n)-\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}-...)=\frac{1}{2}\sqrt{n}
$$
$$
f(n)+ (-\frac{1}{2.2} +\frac{1}{3!})f''(n)+(\frac{1}{2.3!} -\frac{1}{4!})f'''(n)...=\frac{2}{3}n^\frac{3}{2}+\frac{1}{2}\sqrt{n}+c
$$
$$
 f''(n)-\frac{f'''(n)}{2!}+\frac{f^{4}(n)}{3!}-...)=\frac{d(\sqrt{n})}{dn}=\frac{1}{2\sqrt{n}}
$$
If you continue in that way to cancel $f^{r}(n)$ terms step by step,  you will get
$$
f(n)=c+\frac{2}{3}n^\frac{3}{2}+\frac{1}{2}\sqrt{n}+a_2\frac{1}{\sqrt{n}}+a_3\frac{1}{n\sqrt{n}}+a_4\frac{1}{n^2\sqrt{n}}+....
$$
You can find $a_n$ constants by Bernoulli numbers, please see Euler-Maclaurin formula. I just wanted to show the method. http://planetmath.org/eulermaclaurinsummationformula
You can also apply the same method for  $\sum_{i=1}^n(i^k)=P(n)$, $k$ is any real number .
you can get
$$
\sum_{i=1}^n(i^k)=P(n)=c+\frac{1}{k+1}n^{k+1}+\frac{1}{2}n^{k}+b_2kn^{k-1}+....
$$
$$
P(1)=1=c+\frac{1}{k+1}+\frac{1}{2}+b_2k+....
$$
$$
=c=1-\frac{1}{k+1}-\frac{1}{2}-b_2k+....
$$
A: The comparison of this sum with a Riemann integral yields
$$
\frac1{n\sqrt{n}}\sum_{i=1}^n\sqrt{i}=\int_0^1\sqrt{t}\mathrm dt+\sum_{i=1}^n\int_{(i-1)/n}^{i/n}(\sqrt{i/n}-\sqrt{t})\mathrm dt,
$$
that is,
$$
\frac1{n\sqrt{n}}\sum_{i=1}^n\sqrt{i}=\frac23+\frac1{n\sqrt{n}}\sum_{i=1}^n\int_0^1(\sqrt{i}-\sqrt{i-t})\mathrm dt,
$$
or,
$$
\sum_{i=1}^n\sqrt{i}=\frac23n\sqrt{n}+\sum_{i=1}^na_i,\qquad a_i=\int_0^1(\sqrt{i}-\sqrt{i-t})\mathrm dt.
$$
Note that, when $i\to\infty$,
$$
a_i=\int_0^1\frac{t}{\sqrt{i}+\sqrt{i-t}}\mathrm dt\sim\frac1{2\sqrt{i}}\int_0^1t\mathrm dt\sim\frac1{4\sqrt{i}},
$$
and that the series $\sum\limits_i\frac1{4\sqrt{i}}$ diverges, hence
$$
\sum_{i=1}^na_i\sim\sum_{i=1}^n\frac1{4\sqrt{i}}\sim\int_0^n\frac{\mathrm dt}{4\sqrt{t}}=\frac12\sqrt{n}.
$$
Using this equivalent in the formula above, one gets
$$
\sum_{i=1}^n\sqrt{i}=\frac23n\sqrt{n}+\frac12\sqrt{n}+o\left(\sqrt{n}\right)=\frac23n\sqrt{n+\frac32}+o\left(\sqrt{n}\right).
$$
A: 
(Edited by user @Did on 2017, Jan 27) As explained in the comments for 3.5 years now, the correction $n+\frac54$ suggested in this answer is incorrect. For all purposes, this correction should read $n+\frac32$, see my answer for some explanations.


Since you haven't received a response yet, I can address the first bit of the question with a decent approximation,

$$\sum\limits_{i=1}^{n} \sqrt{i} \approx \frac{2}{3}n \sqrt{\frac{5}{4} + n}$$

This approximation isn't surprising, considering the fact that
$$\int\limits_{1}^{n} \sqrt{t} dt = \frac{2}{3} \left( n^{3/2} - 1\right).$$
The data looks like,
$$\newcommand\T{\Rule{0pt}{1em}{.3em}}
\begin{array}{|l|l|}
\hline {\rm n} & \sum\limits_{i=1}^n\sqrt {i} & \tfrac23n\sqrt{\tfrac54+n} \T \\\hline
  1 \T & 1 & 1 \\\hline
  5 \T & 8.382 & 8.333 \\\hline
  10 \T & 22.468 & 22.361 \\\hline
  25 \T & 85.634 & 85.391  \\\hline
  50 \T & 239.04 & 238.63 \\\hline
  100 \T & 671.4 & 670.82 \\\hline
  200 \T & 1892.5 & 1891.5 \\\hline
  500 \T & 7464.5 & 7462.9 \\\hline
  10000 \T & 666716 & 666708 \\\hline
\end{array}$$
A: Entering the query into WolframAlpha (query sum i=1 to n of sqrt(i)) gives two answers:


*

*By the definition of generalized harmonic numbers, the sum is the harmonic number $H_{n,{-\frac{1}{2}}}$

*$-\zeta\left(-\frac{1}{2}, n+1\right)-\frac{\zeta\left(\frac{3}{2}\right)}{4 \pi}$ where $\zeta$ is the Hurwitz zeta function
