I got this question in an exam (high school level) I recently appeared in.
If $ \mathop{\sum}\limits_{{k}{=}{1}}\limits^{80}{\frac{1}{\sqrt{k}}}{=}{S} $, then find the greatest integer less than or equal to $S$.
This is supposed to be a problem from 'sequences and series' so I tried converting the sum into a telescopic sum, but to no avail. The closest I've come to solving it is by taking a completely different approach. I evaluated $ \mathop{\int}\limits_{1}\limits^{80}{\frac{1}{\sqrt{x}}}dx $ to put a lower bound to the sum. This value comes out to be approximately 15.8, but the answer to the problem is 16 so this isn't really helpful. Anyhow, I feel like I am nowhere near solving this question so any help would be greatly appreciated.
Thank you.
If the curve $y=1/\sqrt{x}$ is drawn over the interval $[1,81],$ the sum appears above the curve as a step function, so the sum is at least the value of the integral of $1/\sqrt{x}$ from $1$ to $81,$ which is exactly $16.$
On the other hand, one can see that $S-1,$ (drop the first box) where $S$ is the sum, fits under the graph of $y=1/\sqrt{x}$ over the interval $[0,80],$ so that $S-1 \le 8 \sqrt{5}-2,$ the value of the integral, leading to $S \ge 1+8\sqrt{5}-2=16.88...$
So the sum is definitely at least $16$ and at most about $16.9$ , in agreement with the answer given as $16.$