Asymptotic for the following series. $$S_N=\frac{\sum_{i=1}^{2N}\sqrt{i}+\sqrt{i-1}}{\sum_{i=1}^{N}\sqrt{i}+\sqrt{i-1}}$$
I want to find an equivalence/asymptotic for $S_N$ as $N$ become very large. I tried the following: 
Edit
We know that $$\sum_{i=1}^{N}\sqrt{i}\approx\frac{2N^{3/2}}{3}$$ and so $$S_N\approx\frac{(2N)^{3/2}+(2N-1)^{3/2}}{N^{3/2}+(N-1)^{3/2}}\approx 2\sqrt{2}.$$ Is this estimation correct and also how can it be improved?

Here is a
reasonably elementary proof
for an arbitrary exponent,
not just $\frac12$.
It shows that
if the exponent is $a$
with $a > 0$,
then the limit is
$2^{a+1}$.
It also gives explicit bounds.
It is based on
this:
If $a > 0$ then
$\dfrac{n^{a+1}}{a+1}+n^a
\gt   \sum_{1}^{n} k^a
\gt \dfrac{n^{a+1}}{a+1}
$.
(Similar bounds can be gotten
for $a < 0$.)
Proof:
$(k-1)^a 
< \int_{k-1}^{k} x^a dx
\lt k^a
$
so
$\sum_{k=1}^n (k-1)^a
\lt \sum_{k=1}^n  \int_{k-1}^{k} x^a dx
\lt \sum_{k=1}^nk^a
$
or
$\sum_{k=0}^{n-1} k^a
\lt   \int_{0}^{n} x^adx
\lt  \sum_{1}^{n} k^a
$
so that,
since 
$\int_{0}^{n} x^adx
=\dfrac{n^{a+1}}{a+1}
$,
$-n^a
\lt   \dfrac{n^{a+1}}{a+1}-\sum_{1}^{n} k^a
\lt 0
$.
or
$\dfrac{n^{a+1}}{a+1}+n^a
\gt   \sum_{1}^{n} k^a
\gt \dfrac{n^{a+1}}{a+1}
$.
Let
$p(n)
=\sum_{k=1}^n (k^a+(k-1)^a)
$
and
$S(n)
=\dfrac{p(2n)}{p(n)}
$.
$\begin{array}\\
p(n)
&=\sum_{k=1}^n (k^a+(k-1)^a)\\
&=\sum_{k=1}^n k^a+\sum_{k=1}^n(k-1)^a\\
&=\sum_{k=1}^n k^a+\sum_{k=0}^{n-1}k^a\\
&=\sum_{k=1}^n k^a+\sum_{k=1}^{n-1}k^a\\
&=2\sum_{k=1}^n k^a-n^a\\
\end{array}
$
so that
$2\dfrac{n^{a+1}}{a+1}-n^a
\lt p(n)
\lt 2\dfrac{n^{a+1}}{a+1}+2n^a
$.
Therefore
$\begin{array}\\
S(n)
&\lt \dfrac{2\dfrac{(2n)^{a+1}}{a+1}+2(2n)^a}{2\dfrac{n^{a+1}}{a+1}-n^a}\\
&= \dfrac{ 2^{a+2}+(a+1)2^{a+1}/n}{2-(a+1)/n}\\
&= \dfrac{ 2^{a+1}+(a+1)2^{a}/n}{1-(a+1)/(2n)}\\
\end{array}
$
and
$\begin{array}\\
S(n)
&\gt \dfrac{2\dfrac{(2n)^{a+1}}{a+1}-(2n)^a}{2\dfrac{n^{a+1}}{a+1}+2n^a}\\
&= \dfrac{ 2^{a+2}-2(a+1)2^{a+1}/n}{2+2(a+1)/n}\\
&= \dfrac{ 2^{a+1}-2(a+1)2^{a}/n}{1+2(a+1)/n}\\
\end{array}
$
Therefore
$\lim_{n \to \infty} S(n)
=2^{a+1}
$.
