After recently watching a Numberphile video about a square problem I started thinking about what would happen to the sum of all angles if you had n amount of squares. After a bit of testing, I noticed that:

$$In\space degrees\space and\space assuming\space each\space square\space is\space 1 \times 1\space units$$ $$\left(\sum\limits_{n=1}^{10^{x+1}} arctan\left(\frac{1}{n}\right)\right) - \left(\sum\limits_{n=1}^{10^x} arctan\left(\frac{1}{n}\right)\right) \approx 131.93 $$

So, in words, the sum of the angles of 100 squares stacked next to each other minus the sum of the angles of 10 squares stacked next to each other is approximately equal to 131.93. This is true for any $10^{x+1}$ angles minus $10^x$ angles. I wrote a simple program to show some results:

10^2 angles - 10^1 angles = 129.31207799146
10^3 angles - 10^2 angles = 131.67011361562
10^4 angles - 10^3 angles = 131.90261998016
10^5 angles - 10^4 angles = 131.92582944096
10^6 angles - 10^5 angles = 131.92814996682
10^7 angles - 10^6 angles = 131.92838201514
10^8 angles - 10^7 angles = 131.92840522042

The difference of $10^{x+1}$ angles and $10^x$ angles seems to be approaching some number. Is it possible to find the number that is being approached, and if so, what is that number and how do you find it?


Your limit is $\ln 10$ in radians or $\frac{180\ln 10}\pi$ degrees. The reason is that for large $n$, $\arctan \frac1n\approx \frac1n$ and the $m$th partial sum of the harmonic series is essentially $\ln m$. By cancelling constants and effects of small $n$, your expression is approximately $\ln10^{x+1}-\ln10^x=\ln 10$.

  • $\begingroup$ Oh wow, I didn't think of that, thank you. I'll accept your answer as soon as I can. $\endgroup$ – David Dec 13 '14 at 22:21

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