Geometry problem boils down to finding a closed form for $\sum_{n=1}^{k}{\arctan{\left(\frac{1}{n}\right)}}$

Question

I was solving the following problem:

"Find $\angle A + \angle B + \angle C$ in the figure below, assuming the three shapes are squares." And I found a beautiful one-liner using complex numbers:

$(1+i)(2+i)(3+i)=10i$, so $\angle A + \angle B + \angle C = \frac{\pi}{2}$

Now, I thought, what if I want to generalize? What if, instead of three squares, there were $2018$ squares? What would the sum of the angles be then? Could I make a formula for $k$ squares?

Essentially, the question boiled down to finding a closed form for the argument of the complex number

$$\prod_{n=1}^{k}{(n+i)}=\prod_{n=1}^{k}{\left(\sqrt{n^2+1}\right)e^{i\cot^{-1}{n}}}$$

This we can break into two parts, finding a closed form for $$\prod_{n=1}^{k}{(n^2+1)}$$ and $$\sum_{n=1}^{k}{\cot^{-1}{n}}$$

This I don't know how to solve.

Answer 1

With CAS help:

$$\sum _{n=1}^k \tan ^{-1}\left(\frac{1}{n}\right)=\\\int \left(\sum _{n=1}^k \frac{\partial }{\partial a}\tan ^{-1}\left(\frac{a}{n}\right)\right) \, da=\\\int \left(\sum _{n=1}^k \frac{n}{a^2+n^2}\right) \, da=\\\int \frac{1}{2} \left(-H_{-i a}-H_{i a}+H_{-i a+k}+H_{i a+k}\right) \, da=\\-\frac{1}{2} i (\text{log$\Gamma $}(1-i a)-\text{log$\Gamma $}(1+i a)-\text{log$\Gamma $}(1-i a+k)+\text{log$\Gamma $}(1+i a+k))+C$$

where $a=1$ and $C=0$ then:

$$\color{blue}{\sum _{n=1}^k \tan ^{-1}\left(\frac{1}{n}\right)=-\frac{1}{2} i (\text{log$\Gamma $}(1-i)-\text{log$\Gamma $}(1+i)-\text{log$\Gamma $}((1-i)+k)+\text{log$\Gamma $}((1+i)+k))}$$

Where: $H_{i a}$ is harmonic number and $\text{log$\Gamma $}(1-i)$ is logarithm of the gamma function

MMA code:

HoldForm[Sum[ArcTan[1/n], {n, 1, k}] == -(1/2)
I (LogGamma[1 - I] - LogGamma[1 + I] - LogGamma[(1 - I) + k] + LogGamma[(1 + I) + k])] // TeXForm

Answer 2

We have that $$ \prod\limits_{1\, \le \,n\, \le \,k} {\left( {i + n} \right)} = {1 \over i}\prod\limits_{0\, \le \,n\, \le \,k} {\left( {i + n} \right)} = {1 \over i}i^{\,\overline {\,k + 1\,} } = \left( {1 + i} \right)^{\,\overline {\,k\,} } = {{\Gamma \left( {1 + i + k} \right)} \over {\Gamma \left( {1 + i} \right)}} = k!\left( \matrix{ i + k \cr k \cr} \right) $$ where $x^{\,\overline {\,k\,} } = {{\Gamma (x + k)} \over {\Gamma (x)}}$ denotes the Rising Factorial and $x^{\,\underline {\,k\,} } = \left( {x - k + 1} \right)^{\,\overline {\,k\,} } $ the Falling Factorial.
By means of the expression through The Gamma Function, they are defined as meromorphic functions even for complex $x$ and $k$.

Then $$ \ln \left( {z^{\,\overline {\,k\,} } } \right) = \ln \left( {\left| {z^{\,\overline {\,k\,} } } \right|} \right) + i\arg \left( {z^{\,\overline {\,k\,} } } \right) = \ln {{\Gamma (z + k)} \over {\Gamma (z)}} $$

The above tells us that your question is related to the absolute value and argument of the Gamma Function, which unfortunately do not have a closed expression, better than the above.

Answer 3

Playing around with Wolfy suggests that $s(n) =\sum_{k=1}^n \arctan(1/k) =\frac{3\pi}{4}-\frac12\arctan(g(n)) $ where $g(n)$ is an increasingly complicated fraction.

Some values are $g(4) = 15/8, g(5) = 140/71, g(6) = 2848/7665, g(7) = 14697/203896 $.

To get a recurrence for $g(n)$,

$\begin{array}\\ s(n+1)-s(n) &=\arctan(1/(n+1))\\ &=(\frac{3\pi}{4}-\frac12\arctan(g(n+1)))-(\frac{3\pi}{4}-\frac12\arctan(g(n)))\\ &=\frac12(\arctan(g(n))-\arctan(g(n+1))\\ \end{array} $

so, using $\arctan(x)\pm\arctan(y) =\arctan(\frac{x\pm y}{1\mp xy}) $,

$\begin{array}\\ \arctan(g(n+1)) &=\arctan(g(n))-2\arctan(1/(n+1))\\ &=\arctan(g(n))-\arctan(\frac{2/(n+1)}{1-1/(n+1)^2})\\ &=\arctan(g(n))-\arctan(\frac{2(n+1)}{(n+1)^2-1})\\ &=\arctan(g(n))-\arctan(\frac{2(n+1)}{n^2+2n})\\ &=\arctan(\frac{g(n)-\frac{2(n+1)}{n^2+2n}}{1+g(n)\frac{2(n+1)}{n^2+2n}})\\ &<\arctan(g(n)-\frac{2}{n+1})\\ \end{array} $

so, assuming that the proper branch of arctan is taken, $g(n+1) =\frac{g(n)-\frac{2(n+1)}{n^2+2n}}{1+g(n)\frac{2(n+1)}{n^2+2n}} \lt g(n)-\frac{2}{n+1} $.

Since $\sum 1/n$ diverges, this shows that exentually $g(n) < 0$. At this point the next branch of arctan has to be taken.

For example, Wolfy calculates that $s(20) =\frac{5 π}{4} - \frac12 \arctan(\frac{47183650393321137025}{17864397263976449928}) $ and $s(40) =\frac{5π}{4} + \frac12 \arctan(\frac{41279370979134545450499387615832498927444174194743607}{269197868658553203529942799672226208517623565372926024}) $.

I'll leave it at this.

Answer 4

Note that the addition formula for the arctangent gives $\arctan(x)+\arctan(y)=\arctan(\dfrac{x+y}{1-xy})$; thus, if we define $\alpha_n$ by $\arctan(\alpha_n)=\sum_{i=1}^n\arctan(\frac1i)$, then we have $\alpha_n=\dfrac{\alpha_{n-1}+\frac1n}{1-\frac{\alpha_{n-1}}{n}}$ $=\dfrac{n\alpha_{n-1}+1}{n-\alpha_{n-1}}$; alternately, setting $\alpha_n=\frac{a_n}{b_n}$, we can write this as $a_n=na_{n-1}+b_{n-1}, b_n = -a_{n-1}+nb_{n-1}$ - or $\begin{pmatrix}a_n \\ b_n \end{pmatrix} = \begin{pmatrix}n & 1 \\ -1 & n\end{pmatrix}\begin{pmatrix}a_{n-1} \\ b_{n-1} \end{pmatrix}$. This 'blows up' at $n=3$ (since there's a division by zero), but because the equation in terms of $a_n$ and $b_n$ is homogeneous, we can continue through that point with a rescaling: $(a_n, b_n) (1\leq n\leq 3) =\langle(1, 1), (3, 1), (10, 0)\rangle$, and rescaling to take $a_3=1, b_3=0$ we get $a_4=4, b_4=-1$; $a_5=19, b_5=-9$; $a_6=105, b_6=-73$; $a_7=662, b_7=-616$; etc. Unfortunately, I also can't find any references to this series, and the OEIS doesn't seem to be any help.