Can this upper bound for $\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$ be improved? 
I would like to find the smallest possible upper bound for the following sum of prime radicals (OEIS A062048):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$

This is my attempt. It is too big, I am sure there is a better way of calculating this, but my knowledge is not very good (the questions are at the end of the explanation):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor = \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor$

by Bertrand's postulate $\lfloor \sqrt{p_{i+1}} \rfloor \lt \lfloor \sqrt{2 \cdot p_i} \rfloor$ so it is possible to replace in a first step $\lfloor \sqrt{3} \rfloor$ by $\lfloor \sqrt{2 \cdot 2} \rfloor$:

$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2 \cdot 2} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor$

then recursively doing the same with the next prime $\lfloor \sqrt{5} \rfloor \lt \lfloor \sqrt{2\cdot 3} \rfloor \lt \lfloor \sqrt{2\cdot (2 \cdot 2)} \rfloor$, etc. the result is:

$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_i} \rfloor... \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2^2} \rfloor + \lfloor \sqrt{2^3} \rfloor + ... + \lfloor \sqrt{2^n} \rfloor =$
$=\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}}=$

$\sqrt{2} \lt 2$:

$\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}} \lt 2 \cdot \sum_{i=1}^n 2^{\frac{n}{2}} $

Finally, $\sum_{i=1}^n 2^i = (2^{i+1}-1)-2^0 = 2^{i+1}-2$, thus:

$2 \cdot \sum_{i=1}^n 2^{\frac{n}{2}} \lt 2 \cdot (2^{(\frac{n}{2}+1)}-2)=2^{(\frac{n}{2}+2)}-4$

So finally the expression would be:

$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P \lt 2^{(\frac{n}{2}+2)}-4$
I would like to ask the following questions:


*

*Are the calculations correct or there are gaps / errors?


*Is there a better approach? Thank you!

UPDATE 2016/7/17: Thank you very much for the answers, for the sake of completeness I have sketched the solutions everybody kindly provided, compared with $\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$. The reason is that this sequence is expected to grow slower than the sequence of the sum of the first $n$ elements, so it is a good expression to compare with. The only solution that was too big was mine, kindly improved by mathlove, that is why I wanted to ask for better values really under the value of the sum of the first $n$ natural numbers.
Blue: $\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$
Red (mathlove) = $\sum_{i=1}^{n}\sqrt{p_i}\lt \sum_{i=1}^{n}2^{\frac i2}=\frac{2^{\frac{n+1}{2}}-\sqrt 2}{\sqrt 2-1}$
Yellow (Marco Cantarini) = $n\sqrt{n\log\left(n\right)+n\log\left(\log\left(n\right)\right)}-\frac{2\left(n-1\right)\sqrt{n-1}}{3\sqrt{\log\left(2\right)}}+1$
Green (LeGrandDODOM) (leading term only): $\displaystyle\frac 23 n^{\frac 32}\sqrt{\ln n}$



$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_i} \rfloor... \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2^2} \rfloor + \lfloor \sqrt{2^3} \rfloor + ... + \lfloor \sqrt{2^n} \rfloor =$
  $=\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}}=$

Then, you used $\sqrt{2} \lt 2$, but I think you don't need to do so because
$$\sqrt 2\times 2^{\frac{n-1}{2}}=2^{\frac{n}{2}}$$
so the sum can be written as
$$\sum_{i=1}^{n}2^{\frac i2}\tag1$$
(By the way, it seems that you have
$$\sum_{i=1}^n 2^{\frac{i}{2}} =2^{(\frac{n}{2}+1)}-2$$
but this is incorrect. See when $n=1$.)
Finally, from $(1)$,
$$\sum_{i=1}^{n}\sqrt{p_i}\lt \sum_{i=1}^{n}2^{\frac i2}=\frac{2^{\frac{n+1}{2}}-\sqrt 2}{\sqrt 2-1}$$
