Infinite Trig Sum

The series can be written as

$\displaystyle\sum_{n=1}^{\infty} \cot^{-1}(n^{2} + n + 1) = \sum_{n=1}^{\infty} \tan^{-1} \frac{1}{(n^{2} + n + 1)}$ .

Next, using the fact that

$\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}(\frac{x - y}{1 + xy})$

let $x = n + 1$ and $y = n$ to see that

$\tan^{-1}(n + 1) - \tan^{-1}(n) = \tan^{-1} \dfrac{1}{(n^{2} + n + 1)}$ .

Our series can then be written as

$\displaystyle\sum_{n=1}^{\infty} (\tan^{-1}(n + 1) - \tan^{-1}(n))$ ,

which is a telescoping series, giving a sum of

$-\tan^{-1}(1) + \lim_{n \rightarrow \infty} (\tan^{-1}(n + 1)) = -\dfrac{\pi}{4} + \dfrac{\pi}{2} = \dfrac{\pi}{4}$ ,

which in degrees is $\boxed{45^{\circ}}$ .

First you have to figure out that the $n^{th}$ term is $cot^{-1}(n^2 + n + 1)$ (that should be more clear, by the way).

Recall that the tangent sum formula states that $tan(a+b) = \frac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ . Now let $p = tan(a)$ and $q = tan(b)$ . Then if $-90^o < tan^{-1}(p) + tan^{-1}(q) < 90^o$ , we have $tan^{-1}(p) + tan^{-1}(q) = tan^{-1} \left(\frac{p+q}{1-pq} \right)$ Now if we let $x = \frac{1}{p}$ and $y = \frac{1}{q}$ , then this becomes: $tan^{-1} \left(\frac{1}{x} \right) + tan^{-1} \left( \frac{1}{y} \right) = tan^{-1} \left(\frac{x+y}{xy-1} \right)$ $cot^{-1} (x) + cot^{-1} (y) = cot^{-1} \left(\frac{xy-1}{x+y} \right) (*)$

Using this formula, I calculated the first few partial sums. They turned out to be $cot^{-1} \left(\frac{3}{1}\right), cot^{-1}\left(\frac{4}{2}\right), cot^{-1} \left(\frac{5}{3} \right), ...$

So we can prove by induction that the $n^{th}$ partial sum is equal to $cot^{-1} \left(\frac{n+2}{n}\right)$ . This is very easy using formula $(*)$ above:

The base case $n = 1$ is clearly true since $cot^{-1}(3) = cot^{-1}\left(\frac{1+2}{1}\right)$ .

Now suppose the $k^{th}$ partial sum is $cot^{-1}\left(\frac{k+2}{k}\right)$ . Then the $(k+1)^{th}$ partial sum is: $cot^{-1}\left(\frac{k+2}{k}\right) + cot^{-1}\left((k+1)^2+(k+1)+1\right)$ $= cot^{-1} \left(\frac{\frac{k+2}{k}*(k^2+3k+3) - 1}{\frac{k+2}{k} + k^2+3k+3} \right)$ $= cot^{-1} \left( \frac{k^3+5k^2+9k+6 - k}{k+2 + k^3 + 3k^2+3k}\right)$ $= cot^{-1} \left( \frac{(k+3)(k^2+2k+2)}{(k+1)(k^2+2k+2)} \right) = cot^{-1} \left(\frac{k+3}{k+1} \right)$ Hence the induction is complete.

Therefore, the infinite sum is equal to $\lim_{n \to \infty} cot^{-1} \left(\frac{n+2}{n}\right) = cot^{-1} (1) = \boxed{45^o}$

Let us define $\displaystyle \theta_n = \cot^{-1} 1+\cot^{-1} 3 + \cot^{-1} 7 + \cdots + \cot^{-1} (n^2+n+1)$ . Then $\theta_n = \tan^{-1} (n+1)$ radians for all integers $n\ge 0$ . Let us prove the claim by induction.

For $n=0$ , $\theta_0 = \cot^{-1} 1 = \tan^{-1} (0+1)$ , therefore the claim is true for $n=0$ . Assuming that the claim is true for $n$ , then for $n+1$ :

$\begin{aligned} \theta_{n+1} & = \theta_n + \cot^{-1} ((n+1)^2 + (n+1)+1) \\ & = \tan^{-1} (n+1) + \tan^{-1} \frac 1{(n+1)^2+(n+1)+1} \\ & = \tan^{-1} \left(\frac {n+1+\frac 1{(n+1)^2+(n+1)+1}}{1-\frac {n+1}{(n+1)^2+(n+1)+1}}\right) \\ & = \tan^{-1} \left(\frac {(n+1)^3+(n+1)^2+(n+1)+1}{(n+1)^2+1} \right) \\ & = \tan^{-1} ((n+1)+1) \end{aligned}$

The claim is also true for $n+1$ , therefore it is true for all $n \ge 0$ .

Note that $\theta_n$ includes an additional initial term of $\cot^{-1} 1$ , therefore the answer we need to find is

$\begin{aligned} \lim_{n \to \infty} \theta_n - \cot^{-1} 1 & = \lim_{n \to \infty} \tan^{-1}(n+1) - \frac \pi 4 \\ & = \frac \pi 2 - \frac \pi 4 = \frac \pi 4 = \boxed{45}^\circ \end{aligned}$