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So I was watching an "old" video from numberphile about the three square problem. https://youtu.be/m5evLoL0xwg Here is also an image: http://mathforlove.com/wp-content/uploads/2014/09/Screen-Shot-2014-09-21-at-6.59.05-PM.png

It's pretty easy to see that the sum of three angles is 90°, but now I am curious what if we keep going on. What if we had more than 3 suqars and with that more than 3 angles. What would the sum be? Basically I want to find the value of: $\lim\limits_{n \rightarrow \infty} \sum\limits_{k=1}^n \arctan(\frac{1}{\sqrt{k^2+1}})$

If there is an answer I would like to know it. Thank you. P.S. $\arctan$ stands for $\tan^{-1}$

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  • $\begingroup$ Please give a summary of the relevant information from the video in your post. You can't expect people to watch a 12-minute youTube clip in order to understand your question. $\endgroup$ – John Gowers Jun 20 '15 at 19:25
  • $\begingroup$ As it relates to generalizing the three-square problem, it should really just be $1/k$ inside the arctan, not $1/\sqrt{k^2+1}$. (Note, there is a link to the question here from math.stackexchange.com/questions/2800600/… .) $\endgroup$ – Barry Cipra May 29 '18 at 20:31
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The limit does not exist.

Indeed, for $0\le x< \pi$ we have

\begin{align} \tan(x)=\frac{\sin(x)}{\cos(x)}\le\sin(x)\le x \end{align}

Since $\arctan$ is increasing for the appropriate range of values of $x$, we get:

$$ x\le\arctan (x) $$

for all $x$.

Now we get, for all $k\ge 1$: \begin{align} \arctan\left(\frac{1}{\sqrt{k^2+1}}\right) & \ge\frac1{\sqrt{k^2+1}}\\ &\ge \frac1{\sqrt{k^2+3k^2}}\\ &=\frac{1}{2k} \end{align}

Since the series $\sum_{k=0}^\infty \frac{1}{k}$ is known to diverge, your series diverges too.

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  • $\begingroup$ One of the initial inequalities seems to be backwards: $x\le\tan x$ for $x\in[0,\frac\pi2)$. This can be repaired, though: since arctan is concave on $[0,\infty)$, the secant over $[0,\frac1{\sqrt2}]$ lies under the graph; thus $\arctan x\ge cx$ with $c = \arctan(1/\sqrt2)/(1/\sqrt2)$. The argument can then proceed as before, just with the extra $c$. $\endgroup$ – user21467 Jun 21 '15 at 21:27
  • $\begingroup$ @StevenTaschuk Feel free to edit it so it's correct. $\endgroup$ – John Gowers Jun 21 '15 at 21:50

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