Let $V$ be the space of all polynomials $f : \mathbb{N}_{\ge 0} \to F$ (where $F$ is a field of characteristic zero). Define the forward difference operator $\Delta f(n) = f(n+1) - f(n)$. It is not hard to see that the forward difference of a polynomial of degree $d$ is a polynomial of degree $d-1$, hence defines a linear operator $V_d \to V_{d-1}$ where $V_d$ is the space of polynomials of degree at most $d$. Note that $\dim V_d = d+1$.
We want to think of $\Delta$ as a discrete analogue of the derivative, so it is natural to define the corresponding discrete analogue of the integral $(\int f)(n) = \sum_{k=0}^{n-1} f(k)$. But of course we need to prove that this actually sends polynomials to polynomials. Since $(\int \Delta f)(n) = f(n) - f(0)$ (the "fundamental theorem of discrete calculus"), it suffices to show that the forward difference is surjective as a linear operator $V_d \to V_{d-1}$.
But by the "fundamental theorem," the image of the integral is precisely the subspace of $V_d$ of polynomials such that $f(0) = 0$, so the forward difference and integral define an isomorphism between $V_{d-1}$ and this subspace.
More explicitly, you can observe that $\Delta$ is upper triangular in the standard basis, work by induction, or use the Newton basis $1, n, {n \choose 2}, {n \choose 3}, ...$ for the space of polynomials. In this basis we have $\Delta {n \choose k} = {n \choose k-1}$, and now the result is really obvious.
The method of finite differences provides a fairly clean way to derive a formula for $\sum n^m$ for fixed $m$. In fact, for any polynomial $f(n)$ we have the "discrete Taylor formula"
$$f(n) = \sum_{k \ge 0} \Delta^k f(0) {n \choose k}$$
and it's easy to compute the numbers $\Delta^k f(0)$ using a finite difference table and then to replace ${n \choose k}$ by ${n \choose k+1}$. I wrote a blog post that explains this, but it's getting harder to find; I also explained it in my notes on generating functions.
If you wanted to prove that
$$
\sum_{k=1}^n \frac 1{\sqrt k} \ge \sqrt n,
$$
that I can do. It is clear for $n=1$ (since we have equality then), so that it suffices to verify that
$$
\sum_{k=1}^{n+1} \frac 1{\sqrt k} \ge \sqrt{n+1}
$$
but this is equivalent to
$$
\sum_{k=1}^{n} \frac 1{\sqrt k} + \frac 1{\sqrt{n+1}} \ge \sqrt{n+1} \
$$
and again equivalent to
$$
\sum_{k=1}^n \frac{\sqrt{n+1}}{\sqrt k} + 1 \ge n+1
$$
so we only need to prove the last statement now, using induction hypothesis. Since
$$
\sum_{k=1}^n \frac 1{\sqrt k} \ge \sqrt n,
$$
we have
$$
\sum_{k=1}^n \frac{\sqrt{n+1}}{\sqrt k} \ge \sqrt{n+1}\sqrt{n} \ge \sqrt{n} \sqrt{n} = n.
$$
Adding the $1$'s on both sides we get what we wanted.
Hope that helps,
Best Answer
$$ \sum_{i=1}^n\sqrt{i}=f(n) $$
$$ \sum_{i=1}^{n-1}\sqrt{i}=f(n-1) $$
$$ f(n)-f(n-1)=\sqrt{n} \tag 1$$
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
$$ \sum_{i=1}^n(i^k)=P(n)=c+\frac{1}{k+1}n^{k+1}+\frac{1}{2}n^{k}+b_2kn^{k-1}+.... $$
$$ P(1)=1=c+\frac{1}{k+1}+\frac{1}{2}+b_2k+.... $$
$$ =c=1-\frac{1}{k+1}-\frac{1}{2}-b_2k+.... $$