Land Of Summations

By successive applications of the Generalized Mean Value Theorem, we see that, for any $0 < x \le 1$ , we can find $0 < z < y < x$ such that $\frac{\sqrt{1+x} - 1 - \frac12x}{x^2} \; = \; \frac{\frac{1}{2\sqrt{1+y}} - \frac12}{2y} \; = \; \frac{(1+y)^{-\frac12} - 1}{4y} \; = \; \frac{-\frac12(1+z)^{-\frac32}}{4} \; = \; -\frac{1}{8(1+z)^{\frac32}}$ and hence we deduce that $-\tfrac18x^2 \; \le \; \sqrt{1+x} - 1 - \tfrac12x \; \le \; 0 \hspace{2cm} 0 \le x \le 1$ Thus $\frac{k}{2n^2} - \frac{k^2}{8n^4} \; \le \; \sqrt{1 + \tfrac{k}{n^2}} - 1 \; \le \; \frac{k}{2n^2}$ for any $1 \le k \le n$ , and hence $\frac{n+1}{4n} - \frac{(n+1)(2n+1)}{48n^3} \; = \; \frac{n(n+1)}{4n^2} \; \le \; \sum_{k=1}^n \left(\sqrt{1 + \tfrac{k}{n^2}} - 1 \right) \; \le \; \frac{n(n+1)}{4n^2} \; = \; \frac{n+1}{4n}$ for all $n$ , and hence $\lim_{n \to \infty} \sum_{k=1}^n \left(\sqrt{1 + \tfrac{k}{n^2}} - 1 \right) \; = \; \boxed{\tfrac14}$