How do I calculate a sum $\sqrt{\frac {\hbar{(v)} }{4G}} r$ with respect to a changing $v$? How do I calculate a sum based on the variable $v$ in this expression $\sqrt{\frac{\hbar(v) }{4G}} r$ which changes in increments of 1 with $v_i= 1$ to $v_f = 6.25E34$?
Straight forward it would look something like this:
$$\sqrt{\frac {\hbar{((1)/(6.25E34))} }{4G}} r + \sqrt{\frac {\hbar{((2)/(6.25E34))} }{4G}} r + \sqrt{\frac {\hbar{((3)/(6.25E34))} }{4G}} r +...+ \sqrt{\frac {\hbar{((6.25E34)/(6.25E34))} }{4G}} r = \text{total mass}$$

 A: Both the square root function and the integration operation handle multiplicative constants nicely.  In particular, if you have non-negative real numbers $a,b$ then the identity $\sqrt{ab}=\sqrt a\sqrt b$ holds.  So for your title quantity we can say
$$\sqrt{\frac{\hbar (v)}{4G}}r=r\sqrt{\frac{\hbar}{4G}}\sqrt v=C_0\sqrt v$$
where we will use $C_0$ to refer to that constant portion moving forward.  Then we can write the listed integral as
$$\int_{v_i}^{v_f}C_0\sqrt v\text dv=\frac 23C_0\left(v^{\frac 32}\right|_{v_i}^{v_f}$$
since a multiplicative constant may be moved outside an integral.  Evaluation of the integral result can be addressed separately.
Edit: from the new listed sum of terms, the individual sum term would be expressed more like
$$\sqrt{\frac{\hbar v}{4Gv_f}}r$$
and the proposed sum can be approached with
$$\sqrt{\frac{\hbar}{4Gv_f}}r\sum_{v=1}^{v_f}\sqrt v$$
but this has a similar problem to the original in that if $v_f\to\infty$ then the result of the sum will approach infinity as well.  See this question Conversion of Riemann Sum to Integral with Square Root for a very similar problem and how much of a $dx$ is required to make the sum converge...
Edit 2, corrected:
The area under a square root curve $f(x)=c\sqrt x$ in an interval $[a,b]$ is always $\frac 23c(b^{\frac 32}-a^{\frac 32})$.  Summing over a large set of consecutive integers plugged in for $x$ in this function will closely approximate this result, especially as the number of values increases.  So for the listed value of $v_f$ we should expect that the result of the listed sum will be within the best possible error tolerances of $\frac 23C_0v_f=\frac 23v_fr\sqrt{\frac{\hbar}{4G}}$.
