Clever Approximation #2

Algebra Level 5

$\large \left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}} + \cdots + \frac{1}{\sqrt{992016}}\right\rfloor = \, ?$

Notation: $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 1990.

2 solutions

Boi (보이)
Jun 17, 2017

For those who used excel or other calculators to figure this out,

Let $S=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{4}}+\text{ }\cdots\text{ }\dfrac{1}{\sqrt{992016}}$ .

we can easily see that $\sqrt{k}+\sqrt{k-1}<2\sqrt{k}<\sqrt{k+1}+\sqrt{k}$ .

Take the reciprocal of each side and we get $\sqrt{k+1}-\sqrt{k}<\dfrac{1}{2\sqrt{k}}<\sqrt{k}-\sqrt{k-1}$ .

Substitute $k=2,3,4,\cdots,992016$ and add them up all. Then we get:

$\sqrt{992016}-\frac{3}{2}<\sqrt{992017}-\sqrt{2}<\dfrac{1}{2}(S-1)<\sqrt{992016}-1$

Add $\dfrac{1}{2}$ to all sides.

$\sqrt{992016}-1<\dfrac{1}{2}S<\sqrt{992016}-\dfrac{1}{2}$

Multiply $2$ and use the fact that $\sqrt{992016}=996$ .

$1990<S<1991$

Therefore, $\lfloor S\rfloor=\boxed{1990}$ .

Sam Dave
Jun 23, 2017

Integrate one divided by underoot x ×dx from limits ranging fron 1 to 992016

This approximates the sum, but does not give the exact answer. The solution given is 1990, but in reality it must have been an irrational number. Integrating is a good way to approximate these sums, so you get credit for that, but it was only luck that your solution was so close to the correct answer.

Alex Li - 3 years, 11 months ago

i agree bro,this method works for large numbers like this one,but fails when it comes to smaller ones,like it wont work for 1 to 80,but it is effective for four digit number and so on

sam dave - 3 years, 11 months ago

Clever Approximation #2

The answer is 1990.

2 solutions

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