# Olympiad problem -- Sum involving many square roots...

√(2-√(2^(2)-1))+√(4-√(4^(2)-1))+√(6-√(6^(2)-1))+...+√(80-√(80^(2)-1))
How the find it's value

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Nidum
Gold Member
Do you have any ideas yourself ?

Just out of curiosity are there any rules about how these problems have to be solved ? Do the solutions always have to be analytic ones ?

Personally if I wanted an answer to this problem for some practical purpose I would just write a program to sum all the terms numerically - it would only be a few lines of coding .

Well you are not supposed to use a calculator and the answer should be rounded of to nearest integer if needed.

Well i have an idea and here it is:
So if y=2-√(2^(2)-1)
y=2-√3
y(2+√3)=1
And this works for all the terms.

But I don't know how to connect it to the equation

mfb
Mentor
I moved the thread to our homework section.

y(2+√3)=1
Or y=1/(2+√3).
If you look at larger values like 1/(78+√(782-1)) or 1/(80+√(802-1)), do you see how you could approximate that?
Does this lead to a familiar series?

Convering series? But how can I estimate the final and?

mfb
Mentor
With an approximation for this series (well, the finite partial sum, to be more precise).

Can I say this

40
Summation (√2n-√(2n)^(2)-1))
n=1
And how can I find the partial sum for this

mfb
Mentor
Can I say this
This is just your original problem written in a different way. Sure.
And how can I find the partial sum for this
Find a suitable approximation, see above.

I'm posting my solution because this thread wasn't originally in homework section and doesn't appear to be homework.
After some algebra I write the problem as$$\sqrt{2}\sum_{n=1}^{40}\sqrt{n-\sqrt{n^2-.25}}$$
I multiply the sum by,
$$\frac{\sqrt{n+\sqrt{n^2-.25}}}{\sqrt{n+\sqrt{n^2-.25}}}$$
I approximate$$\sqrt{n^2-.25}\approx n$$
And after a little more algebra the sum becomes approximately$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}}$$
I approximate ##\sqrt{2}\approx 1.5 ## ## \sqrt{3}\approx 2## etc. I note the first two terms ##\approx \frac{1.75}{2}##, the next five terms##\approx 5\frac{1}{2}\frac{1}{2}##, the next five terms ##5\frac{1}{2}\frac{1}{3}##, the next ten terms ##\approx 10\frac{1}{2}\frac{1}{4}##, and the last eighteen terms ##\approx 18\frac{1}{2}\frac{1}{6}##. Adding it all up I get 5.2.

fresh_42
Mentor
Adding it all up I get 5.2
Hmm, the program (formula from post #9) ended with ##5.65685 ...##

scottdave
mfb
Mentor
Quite good for an approximation. If we round we are off by 1.

Replace the sum by an integral:
$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx \frac 1 2 \int_{0.7}^{40.5} \frac{dx}{\sqrt{x}} = [\sqrt{x}]_{0.7}^{40} = \sqrt{40}-\sqrt{0.7} \approx 6.3 - 0.85 = 5.45$$
0.7 is black magic based on intuition, the square roots were estimated without calculator.

Compare it to the precise value of the sum:
$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx 5.6338$$

If we round that, we are still off by 1, but now the estimate is within 0.2 of the precise value of the original problem.

We get a better result with a numerical evaluation of the first two terms:
$$\frac{1}{2}\sum_{n=1}^{40}\frac{1}{\sqrt{n}} \approx \frac 1 2 \left( 1+\frac{1}{\sqrt 2} + \int_{2.5}^{40.5} \frac{dx}{\sqrt{x}} \right) = 0.5 + 0.35 + \sqrt{40}-\sqrt{2.5} \approx 0.85 + 6.33 - 1.59 = 5.59$$
Again without calculator (and without using the result from above...), but with a bit more thought about the square roots.

scottdave