How to approximate this series? How to approximate this series, non-numerically?
$ S_n = \sum_{n=1}^{50} \sqrt{n}$
 A: 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 $N = 50$.
A: 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:
$$0\leq 2\left((n+1/2)\sqrt{n+1/2}-(n-1/2)\sqrt{n-1/2}\right)-3\sqrt{n}\leq \frac{1}{16}\left(\frac{1}{\sqrt{n+1/2}}-\frac{1}{\sqrt{n-1/2}}\right).$$
