Epsilon-delta proof for a sum 
Consider $\displaystyle \begin{array}{ccccc}
f & : & \mathbb R^+ & \to & \mathbb R \\
 & & x & \mapsto & \frac{x}{\sqrt{1+x}} \\
\end{array}$
Prove that $\displaystyle \sum_{k=1}^n f\left(\frac{k}{n^2}\right)\to \frac{1}2$

The main trouble here is that $n$ is also inside the sum. My teacher gave a proof that I really don't like, and I'm looking for a $\delta-\epsilon$ or $N-\epsilon$ proof.
Heuristically , $\displaystyle f\left(\frac{k}{n^2}\right) \sim\frac{k}{n^2}$ and summing gives $\frac{1}{2}$.
My attempt for the proof fails because, when doing my inequalities, I get terms that go to infinity on the RHS.
 A: 
Claim $$S_n=\sum_{k=0}^nf(\frac{k}{n^2})\rightarrow \frac{f'(0)}{2}$$ as $n$ tends to $+\infty$

Proof.
$f$ is differentiable at $0$, then $f(x)=f(0)+xf'(0)+o(x)=xf'(0)+o(x)$
Let $\varepsilon >0$,
There exist $$\eta >0, \forall x \in [0,\eta), \qquad \vert f(x)-xf'(0)\vert \le \varepsilon\cdot x$$
Let $N \in \mathbb{N}$ such that $\frac{1}{N}<\eta$. For all $n \ge N$ and for all integer $k$ satisfying $0 \le k \le n$ we have 
$$
0\le \frac{k}{n^2} \le \frac{n}{n^2}<\eta
$$
Then 
$$
\forall n\ge N, \forall k, 0\le k \le n \left\vert f(\frac{k}{n^2})-\frac{k}{n^2}f'(0) \right\vert  \le \varepsilon\frac{k}{n^2}
$$
Now we have,
$$
 \left\vert \sum_{k=0}^{n} f(\frac{k}{n^2})-f'(0)\sum_{k=0}^{n}\frac{k}{n^2} \right\vert
\le \varepsilon \sum_{k=0}^{n}\frac{k}{n^2}$$
Therefore,
$$
\left\vert S_n -f'(0)\frac{1}{2}-f'(0)\frac{1}{2n}\right\vert \le\varepsilon\big(\frac{1}{2}+\frac{1}{2n}\big)\le\varepsilon 
$$
Finally,
$$
\left\vert S_n -f'(0)\frac{1}{2}\right\vert \le \left\vert S_n-\frac{f'(0)}{2}-\frac{f'(0)}{2n}\right\vert+ \left\vert \frac{f'(0)}{2n}\right\vert \le\varepsilon+\frac{\vert f'(0)\vert}{2n}
$$
Now choose  $N_1\ge N$ such that $\frac{\vert f'(0)\vert}{2N_1}<\varepsilon$, thus $\vert S_n-f'(0)/2 \vert< 2\varepsilon$ for all $n\ge N_1$.

$(S_n)$ converges to $\frac{f'(0)}{2}$

