Evaluate the following series involving two square roots

Asked Mar 26 '23 at 07:40

Active Mar 30 '23 at 20:37

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Evaluate $$\sum_{n=1}^{119}\sqrt{2-\frac{1}{n^2+\sqrt{n^4+\frac{1}{4}}}}$$

I first decided to do some simplifications. I rewrote $n^4+\frac14$ as $\left(n^2+n+\frac12\right)\left(n^2-n+\frac12\right)$ But I cannot get things out of the square root. Any hints or solutions are greatly appreciated.

edited Mar 30 '23 at 20:37

Arctic Char

16,007

asked Mar 26 '23 at 07:40

MathStackexchangeIsNotSoBad

4,716

It is a simple whole number – Claude Leibovici Mar 26 '23 at 08:22
Rationalize the denominator. Then study the section on two nested square roots. You should end with a difference between square roots that telescopes when you sum the terms. – Oscar Lanzi Mar 26 '23 at 08:37
@OscarLanzi I ended up with $$\sqrt{2+4n^2-\sqrt{16n^4+4}}$$ what to do now – MathStackexchangeIsNotSoBad Mar 26 '23 at 09:15
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$$ \sqrt {2 + 4n^2 - \sqrt {16n^4 + 4} } = \sqrt {n^2 + (n + 1)^2 } - \sqrt {(n - 1)^2 + n^2 } $$ using the very last equation of this section: https://en.wikipedia.org/wiki/Nested_radical#Two_nested_square_roots – Gary Mar 26 '23 at 09:21
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Does this answer your question? Prove that sum is an integer. – found with Approach0 – Martin R Mar 26 '23 at 09:25
@Gary I saw the last section and also the technique...but I'm not able to fully understand it...pls can you tell me how you did this with steps? – MathStackexchangeIsNotSoBad Mar 26 '23 at 09:27
@MartinR let me see – MathStackexchangeIsNotSoBad Mar 26 '23 at 09:28
@MathStackexchangeIsNotSoBad Well, it says $$ \sqrt {a - \sqrt c } = \sqrt {\frac{{a + \sqrt {a^2 - c} }}{2}} - \sqrt {\frac{{a - \sqrt {a^2 - c} }}{2}} , $$ so... – Gary Mar 26 '23 at 09:35

Evaluate the following series involving two square roots

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