Evaluate $\sum_{n\geq 0} \mbox{arccot}(n^2 + n + 1)$ (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.)
 A: 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
$\frac{\pi}{4}+\arctan(\frac{1}{1})-\arctan(\frac{1}{2})+\arctan(\frac{1}{2})-\arctan(\frac{1}{3})+\arctan(\frac{1}{3})-\arctan(\frac{1}{4})+...=\frac{\pi}{2}.$
A: $\newcommand{\ac}{\text{arccot}}$
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).

*

*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 $\displaystyle{\sum _ {n=1}^{\infty}\ac(n^2+n+1)}.$


*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.


*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.
