Question on computing $\sum_{n=1}^\infty \tan^{-1} \left(\frac{3}{n^2 + n - 1}\right)$

Question

Find $$\sum_{n=1}^\infty \tan^{-1} \left(\frac{3}{n^2 + n - 1}\right)$$

I know this can be simplified to a telescoping series of the form

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

The correct answer, on evaluating the limit for this would be,

$\tan^{-1} (n+2) + \tan^{-1} (n+1) +\tan^{-1} (n)- \tan^{-1} 0 - \tan^{-1} (1) - \tan^{-1} (2) =\frac{ 3\pi}4 - \cot^{-1} 2$

But when summing to infinity why do we have remanent $\tan^{-1} (n+2) + \tan^{-1} (n+1) +\tan^{-1} (n)$ since for every $ \tan^{-1} n$ we would have a $-\tan^{-1} n$. Something similar to the argument how set of natural numbers and integers is equal $(n(N) = n(I))$.

In problems I've done before the telescoping sum usually tends to 0 so I might be thinking of it in the wrong way.

Answer 1

What you did is the partial sum

$$\begin{align}S_n&=\sum_{k=1}^n \tan^{-1} \left(\frac{3}{k^2 + k - 1} \right)\\ \\ S_n&=\tan^{-1} (n+2) + \tan^{-1} (n+1) +\tan^{-1} (n)- \tan^{-1} 0 - \tan^{-1} (1) - \tan^{-1} (2)\end{align}$$

the series equals the limit of the partial sum, namely

$$\sum_{k=1}^\infty \tan^{-1} \left(\frac{3}{k^2 + k - 1} \right)=\lim_{n\to\infty}S_n=\frac{5\pi}4 - \tan^{-1} (2)$$

Answer 2

Using $$ \tan \left(\tan ^{-1} a-\tan ^{-1} b\right)=\frac{a-b}{1+a b}\Rightarrow \tan ^{-1} a-\tan ^{-1} b=\tan ^{-1} \left(\frac{a-b}{1+a b}\right), $$ we have $$ a-b=3 \text { and } 1+a b=n^2+n-1 \Rightarrow a=n+2 \textrm{ and } b=n-1. $$ Hence $$ \tan ^{-1}\left(\frac{3}{n^2+n-1}\right)=\tan ^{-1}(n+2)-\tan ^{-1}(n-1) $$ $$\begin{aligned} S&=\sum_{n=0}^{\infty} \tan ^{-1}\left(\frac{3}{n^2+n-1}\right)\\&=\lim _{N \rightarrow \infty} \sum_{n=0}^N\left[\tan ^{-1}(n+2)-\tan^{-1} (n-1)\right] \\&= \lim _{N \rightarrow \infty}\left[\sum_{n=2}^{N+2} \tan ^{-1} n -\sum_{n=-1}^{N-1} \tan ^{-1} n\right]\\&= \lim _{N \rightarrow \alpha}\left[\tan ^{-1}(N+2)+\tan ^{-1}(N+1)+\tan ^{-1} N\right] -\tan ^{-1}(-1)-\tan ^{-1}(0)-\tan ^{-1} 1\\&= \lim _{N \rightarrow \infty}\left[\tan ^{-1}(N+2)+\tan ^{-1}(N+1)+\tan ^{-1} N\right]\end{aligned} $$ Using the identity twice $$\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} \left(\frac{a+b}{1-a b}\right),$$ we have

$$S= \tan ^{-1} \frac{2 N+3}{1-(N+2)(N+1)}+\tan ^{-1} N =\lim _{N \rightarrow \infty} \tan ^{-1}\left(\frac{N-\frac{2 N+3}{N^2+3 N+1}}{1+\frac{2 N^2+3 N}{N^2+3 N+1}}\right)=\frac{\pi}{2}$$

Answer 3

By using the identity $$\tan^{-1}\left(\frac{3}{n^2+n-1}\right)=\tan^{-1}(n+2)-\tan^{-1}(n-1)$$ for $n>0$, the $n$-th partial sum of the series $$S= \sum_{n=1}^{\infty}\tan^{-1}\left(\frac{3}{n^2+n-1}\right)$$ is $$S_n=\tan^{-1}(n+2)+\tan^{-1}(n+1)+\tan^{-1}(n) -\tan^{-1}(2)-\tan^{-1}(1) -\tan^{-1}(0).$$ Letting $n\rightarrow\infty$ we have $$S=\frac{3\pi}{2}-\tan^{-1}2-\frac{\pi}{4}=\frac{5\pi}{4}-\tan^{-1}2.$$