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Prove that $$\frac{\pi}{4}\le\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}\right)$$

EDIT: inspired by Michael Hardy's suggestion I got that

$$\arcsin \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}=\arcsin\frac{1}{\sqrt{n+1}}-\arcsin\frac{1}{\sqrt{n+2}}$$ and then $$\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}\right)\ge\sum_{n=1}^{\infty} \arcsin\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}\right)\rightarrow\frac{\pi}{4}$$ because $\sum_{n=1}^{\infty} \left(\arcsin\frac{1}{\sqrt{n+1}}-\arcsin\frac{1}{\sqrt{n+2}}\right)=\arcsin \frac{\sqrt{2}}{2}=\frac{\pi}{4}$

Sis & Chris.

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We have \begin{align} \arcsin \left(\dfrac{\sqrt{n+1} - \sqrt{n}}{n+1}\right) & \geq \dfrac{\sqrt{n+1} - \sqrt{n}}{n+1} = \dfrac1{n+1} \cdot \dfrac1{\left(\sqrt{n+1} + \sqrt{n}\right)}\\ & \geq \dfrac1{2(n+1)^{3/2}} \end{align} Hence, $$\sum_{n=1}^{\infty} \dfrac1{2(n+1)^{3/2}} = \dfrac{\zeta(3/2)-1}2$$ Hence, it is sufficient to prove that $$\zeta3/2) \geq 1 + \dfrac{\pi}2$$ –  user17762 Feb 16 '13 at 19:33
    
@Marvis: a small difference there $\approx 0.041579$. On the other hand, since it's a contest math question I think there should be a trick that would give us that $\frac{\pi}{4}$ (it's more a guess). –  Chris's sis Feb 16 '13 at 19:37
    
@Marvis: thanks for your point! –  Chris's sis Feb 16 '13 at 19:48
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I'm guessing the identity $\arcsin a + \arcsin b$ $= \arcsin\left(a\sqrt{1-b^2} + b\sqrt{1-a^2}\right)$ may be relevant. –  Michael Hardy Feb 16 '13 at 19:55
    
@MichaelHardy I was trying to turn that one around (expand $\arcsin(x+y)$ into $t + u$) but it got ugly. (I tried to use $\sin(\arctan(x)) = \frac{x}{\sqrt{x^2 +1}}$ and arctan addition.) –  Chris Feb 16 '13 at 20:01

1 Answer 1

up vote 10 down vote accepted

Since $\arcsin x\ge \arctan x$ for $x \in [0,1]$, thus we shall have $$\arcsin(\frac{\sqrt{n+1}-\sqrt{n}}{n+1})\ge \arctan(\frac{\sqrt{n+1}-\sqrt{n}}{n+1})\ge \arctan\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n}+1} $$ $$= \arctan{\sqrt{n+1}}-\arctan{\sqrt{n}}.$$

(The last equality uses $\arctan(x) - \arctan(y) = \arctan(\frac{x-y}{1 + xy})$.)

Done.

P.S.

$\sum \arcsin(\frac{\sqrt{n+1}-\sqrt{n}}{n+1}) \ge \sum(\arctan{\sqrt{n+1}}-\arctan{\sqrt{n}}) = \pi/2-\arctan(1)=\pi/4$

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@PeterTamaroff He may have a point. In the interval $\[0, 1]$, we have $\sin(\arctan(x)) \le \sin(\arcsin(x))$, since the former is $\frac{x}{\sqrt{x^2 +1}}$, while the latter is simply $x$. Since $\sin$ is increasing on that interval, we obtain $\arctan(x) \le \arcsin(x)$. (Notice that the arguments of $\arcsin$ for all terms in our series lie in $\[0, 1]$. Thus, the inequality is valid for every term in our series.) –  Chris Feb 16 '13 at 20:35
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No, no worries. I find this really clever. (That's why I was so picky.) –  Chris Feb 16 '13 at 20:43
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@Yimin: that's a clever solution. Thanks! (+1) (Maybe you may improve a bit your text such that the proof be clear) –  Chris's sis Feb 16 '13 at 20:50
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+1 good job. It was clever to shift from arcsin to arctan –  Amr Feb 16 '13 at 21:08
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+1 good answer. –  Mhenni Benghorbal Feb 16 '13 at 21:15

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