# Evaluating the floor of a tough looking summation

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.

• The upper and lower bounds on the integral need to be adjusted by $\pm 1$ in order to get lower/upper bounds on such a sum of a decreasing function. Jan 29, 2017 at 18:01
• @coffeemath, I get what you're saying, but I have stated in the post itself that this was supposed to be a sequences and series problem. Also, since the upper bound has an integral part of 16 and the lower bound has an integral part of 15, there would be confusion in deciding whether the actual sum has an integral part of 15 or 16. Thanks for responding anyway. Jan 29, 2017 at 18:04

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 $16.$
• For a simpler example suppose the sum is for $k$ from $1$ to $3,$ so it is $S=1/\sqrt{1}+1/\sqrt{2}+1/\sqrt{3}.$ Then one uses three intervals: $[1,2],[2,3],[3,4]$ and looks at the graph of $1/\sqrt{x}$ over each of these. At the three left endpoints $1,2,3$ of these intervals, $1/\sqrt{x}$ evaluates to three terms whose sum is $S.$ And the horizontal line segments of length $1$ drawn from over the left endpoints and going over each of the three intervals are all above the graph of $1/\sqrt{x}$ over the interval $[1,4].$ Note that this last interval has length 3, the number of terms of $S.$ Jan 31, 2017 at 16:09
• @AnindyaMahajan When I first did such estimates, I had to draw several diagrams to see that one should always have the length of the integration interval equal to the number of summed terms, even if this means going beyond by one step. For getting the upper bound on the sum I had to subtract 1 first, as noted, so then had only 79 terms and $[1,80]$ has length 79. Jan 31, 2017 at 16:27
• Only if, as in your sum, the first term is $1$ you can say $S-1 \le \int_1^m f(x)dx$ provided $f$ is decreasing on $[1,m].$ After subtracting the 1, the terms now start at $1/\sqrt{2}$ and heights of these go with the $79$ right endpoints of intervals $[k,k+1]$ for $k$ from $1$ to $79.$ Jan 31, 2017 at 17:00