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$$\left\lfloor\sum_{r=1}^{80}\int_0^1x^{\sqrt r-1}dx\right\rfloor$$

My try: $$K=\left\lfloor\sum_{r=1}^{80}\int_0^1x^{\sqrt r-1}dx\right\rfloor=\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt r}\right\rfloor=\left\lfloor\frac1{\sqrt 1}+\frac1{\sqrt 2}+\ldots+\frac1{\sqrt{80}}\right\rfloor$$ I can only say: $$\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt {80}}\right\rfloor<K<\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt 1}\right\rfloor\\ \lfloor\sqrt{80}\rfloor<K<\lfloor80\rfloor\\ 8<K<80$$ So I don't know how to find K?

RE60K
  • 17,716

2 Answers2

5

Hint: You need tighter bounds on $\dfrac{1}{\sqrt{r}}$.

Note that $\sqrt{r+1} - \sqrt{r} = \dfrac{1}{\sqrt{r+1}+\sqrt{r}} < \dfrac{1}{2\sqrt{r}} < \dfrac{1}{\sqrt{r}+\sqrt{r-1}} = \sqrt{r} - \sqrt{r-1}$.

Hence, $2(\sqrt{r+1} - \sqrt{r}) < \dfrac{1}{\sqrt{r}} < 2(\sqrt{r} - \sqrt{r-1})$.

Using these bounds will give you two telescoping sums, which are easy to evaluate.

JimmyK4542
  • 54,331
5

Since $\frac{1}{\sqrt{x}}$ is decreasing, we have

$$\int_2^{81} \frac{1}{\sqrt{x}} \,dx < \sum_{k = 1}^{80} \frac{1}{\sqrt{k}} < \int_1^{81} \frac{1}{\sqrt{x}} \,dx,$$

and evaluating gives

$$2 (9 - \sqrt{2}) < \sum_{k = 1}^{80} \frac{1}{\sqrt{k}} < 16.$$

Now, $8 < 9$ implies that $\sqrt{2} < \frac{3}{2}$, so we may weaken the first inequality to $$15 = 2 \left(9 - \frac{3}{2}\right) < \sum_{k = 1}^{80} \frac{1}{\sqrt{k}},$$ and so $$\left\lfloor \sum_{k = 1}^{80} \frac{1}{\sqrt{k}} \right\rfloor = 15.$$

Travis Willse
  • 99,363
  • Oh something like Lower sum and upper sum, this is simpler but I wish not deaccepting answers for it might be hurting to some, yours is perfectly acceptable. – RE60K Oct 16 '14 at 08:05
  • Aditya A way to avoid these situations is to wait more than 6 minutes after an answer is posted, to accept it. – Did Oct 16 '14 at 08:24
  • @Aditya No worries, the important thing is that you understand the problem better. I'd opine that JimmyK4542's solution is actually more elementary, as it doesn't involve calculus. – Travis Willse Oct 16 '14 at 08:26
  • @Did thnx a good suggestion – RE60K Oct 16 '14 at 08:32