How to approximate this series?

Question

How to approximate this series, non-numerically?

$ S_n = \sum_{n=1}^{50} \sqrt{n}$

Answer 1

For a very simple approximation, which is the first step to the Euler-McLaurin formula, use this $f'(n) \sim f(n)-f(n-1) $ so $f(n) \sim \int_{n-1}^{n} f(x) dx $ so $\sum_{n=1}^N f(n) \sim \int_0^{N} f(x) dx $.

(Actually, $f(n) \sim \int_{n-1/2}^{n+1/2} f(x) dx $ is more accurate, but this is an approximation, and definitely not the best.)

Letting $f(n) =\sqrt{n}$, $\sum_{n=1}^N \sqrt{n} \sim \int_0^{N} \sqrt{x} dx = \frac{x^{3/2}}{3/2}\big |_0^N =\frac23 N^{3/2} $.

Then, let $N = 50$.

Answer 2

Consider that: $$2\left((n+1)\sqrt{n+1}-n\sqrt{n}\right)=3\sqrt{n+1}+\frac{n-1-\sqrt{n(n+1)}}{\sqrt{n}+\sqrt{n+1}}.$$ The last term is negative, but greater than $$-\frac{3}{2(\sqrt{n}+\sqrt{n+1})}$$ by the AM-GM inequality. This gives: $$\sum_{n=1}^{50}\sqrt{n}\geq\frac{2}{3}\sum_{n=1}^{50}\left(n\sqrt{n}-(n-1)\sqrt{n-1}\right)=\frac{2}{3}50\sqrt{50},$$ $$-\frac{2}{3}50\sqrt{50}+\sum_{n=1}^{50}\sqrt{n}\leq\frac{1}{2}\sum_{n=1}^{50}\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{1}{2}\sum_{n=1}^{50}\left(\sqrt{n}-\sqrt{n-1}\right)=\frac{1}{2}\sqrt{50}.$$ The same argument proves that the sum $$\sum_{n=1}^{N}\sqrt{n}$$ is always between $$\frac{2}{3}N\sqrt{N}$$ and $$\frac{4N+3}{6}\sqrt{N}.$$ As pointed out in the previous comments, we can produce a tighter bound by considering that: $$0\leq 2\left((n+1/2)\sqrt{n+1/2}-(n-1/2)\sqrt{n-1/2}\right)-3\sqrt{n}\leq \frac{1}{16}\left(\frac{1}{\sqrt{n+1/2}}-\frac{1}{\sqrt{n-1/2}}\right).$$