Evaluate $\sum_{n\geq 0} \mbox{arccot}(n^2 + n + 1)$

Question

(This is a 1986 Putnam Challenge problem.)

First, note that \begin{equation} n^2 + n + 1 = \frac{n^3 - 1}{n - 1}, \end{equation} which is the slope of the secant line through $f(x) = x^3$ at $x = 1$ and $x = n$. So $\mbox{arccot}\Big(\frac{n^3 - 1}{n - 1}\Big)$ is the acute angle that this line makes with any vertical line. Thus the requested sum is the sum of all these angles ... and this is where I get stuck.

Surely I'm not meant to use a trig addition formula here; all the ones I'm aware of are too convoluted for the purpose.

Am I at least on the right track? The form of the polynomial seems to suggest so, but maybe it's a red herring. Do they do that with Putnam problems?

(It might also be relevant that $\mbox{arccot}(x)$ is convex over the positive reals. But that merely provides inequalities, not identities ... or so it would seem.)

Answer 1

We observe that $\operatorname{arccot}(n^2+n+1)=\arctan(\frac{1}{n})-\arctan(\frac{1}{n+1})$ for $n>0$.

Hence, it is telescoping series, adding to

$\frac{\pi}{4}+\arctan(\frac{1}{1})-\arctan(\frac{1}{2})+\arctan(\frac{1}{2})-\arctan(\frac{1}{3})+\arctan(\frac{1}{3})-\arctan(\frac{1}{4})+...=\frac{\pi}{2}.$

Answer 2

$\newcommand{\ac}{\text{arccot}}$

I used to give this problem as a challenge to second-year calculus students. It's kind of nice because my presentation was 'backwards': start with the closed-form, prove it using induction and angle-addition, and then as a real challenge, prove the angle-addition law (nice exercises).

1. I claim that the partial sums have the following formula: $$ \sum _{n=1}^N \ac(n^2+n+1) = \ac\left(\frac{N+2}{N}\right). $$ Assuming this is true, find $\displaystyle{\sum _ {n=1}^{\infty}\ac(n^2+n+1)}.$

2. Recall the angle-addition formula: for $x,y \geq 1,$ $$\ac(x)+\ac(y)=\ac\left(\frac{xy-1}{x+y}\right).$$ Using this, prove the partial-sum formula by induction.

3. Now to prove the angle-addition formula. There are two methods I'd recommend. One way is to observe $x,y$ are interchangeable, verify the identity for $(x,y)=(1,1)$, and then verify the partials with respect to $x$ are equal. Another way is to use geometry or linear algebra to interpret the result as slopes triangles.