The approximation of a summation

In a recent Olympiad contest, a problem reduced to finding the greatest integer function of the summation of the reciprocals of square roots from 1 to 100.

$$\left\lfloor \sum_{k=1}^{100}\frac{1}{\sqrt{k}}\right\rfloor$$

Using some Cauchy, I showed that it was greater than 14 , but Wolfram Alpha says the answer is 18, which is quite far off. Is there any way to do it with elementary math? Even a calculus proof would be fine.

• Did you try approximating with integral $\int _1 ^{100} \frac {1}{\sqrt {x}}dx$ – Archis Welankar May 25 '17 at 12:06
• That just gives 20. – Anish Hebbar May 25 '17 at 12:18
• What its $2\sqrt {x}=2 (10-1)=18$ – Archis Welankar May 25 '17 at 12:51

We have that $$\sqrt k-\sqrt{k-1}=\frac{1}{\sqrt k+\sqrt{k-1}}>\frac{1}{2\sqrt k}=\frac{1}{\sqrt k+\sqrt{k}}>\frac{1}{\sqrt k+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$$ And $$\sum_{k=1}^{100}\sqrt{k+1}-\sqrt k=\sqrt{101}-1>9$$ So we have that $$\sum_{k=1}^{100}\frac{1}{2\sqrt{k}}>9\implies \sum_{k=1}^{100}\frac{1}{\sqrt k}>18$$ We also have that $$\sum_{k=2}^{100}\sqrt k-\sqrt{k-1}=9$$ So we have that $$\sum_{k=2}^{100}\frac1{2\sqrt{k}}<9\implies \sum_{k=2}^{100}\frac{1}{\sqrt k}<18$$ So we have that $$1+\sum_{k=2}^{100}\frac{1}{\sqrt{k}}=\sum_{k=1}^{100}\frac{1}{\sqrt{k}}<19$$

• @Anish hey if you don't understand my answer or need more clarification feel free to comment, otherwise if my answer answered your question consider accepting it. – kingW3 May 26 '17 at 8:54

The sum can be written as:

$$1+\frac{1}{\sqrt2} + \frac1{\sqrt3} +\cdots+\frac 1{\sqrt{100}} = 1+ \frac{\sqrt2}2 +\frac{\sqrt3}3 +\cdots + \frac 1{10}$$

upon rationalising the denominator.

• And how does that help? – orlp May 25 '17 at 12:19
• Graph the sum, and then you calculate the slope, which will give the answer – user373868 May 25 '17 at 12:27
• For the math olympiad you only have pen and paper. Will you really make a graph of size $n = 100$, which is then acccurate enough to use it to calculate the slope? Also how would the slope give the answer? – orlp May 25 '17 at 12:29