Is it possible to do such a sum?
$$\sum_{i=1}^n \sqrt (i^2+\frac{8i}{n}+\frac{16+n^2}{n^2})$$
I want to reach to a function with only the n as variable. I believe that it is possible somehow by squaring the summation and squaring the other side of the equation.
For more specification, this summation is a part of:
$$C = \lim_{n \to ∞} [\frac{2}{n}\sum_{i=1}^n \sqrt (i^2+\frac{8i}{n}+\frac{16}{n^2}+1)]$$
Computing the sum is not needed to determine the value of $C$. To do the latter, note that $$i^2+\frac{8i}n+\frac{16}{n^2}+1\geqslant i^2$$ hence $$\frac2n\sum_{i=1}^n\sqrt{ i^2+\frac{8i}n+\frac{16+n^2}{n^2}}\geqslant\frac2n\sum_{i=1}^ni=n+1$$ hence $$C=+\infty$$ Edit: On the other hand, $$\frac2n\sum_{i=1}^n\sqrt{ 1+\frac{4i^2}{n^2}}$$ is the $n$th Riemann sum of the function $f(x)=\sqrt{1+x^2}$ on $[0,2]$ hence its limit when $n\to\infty$ is direct but there is no reason to expect a closed form for these sums for every finite $n$.