summation of infinite terms contradiction

Question

Consider the sum $$\sum_{i=0}^\infty (\tan^{-1}{(i+1)} - \tan^{-1}(i))$$ if we write $n^{th}$ terms ($T_n$) $$\begin{align} T_0 &= \tan^{-1}(1) - \tan^{-1}(0)\\ T_1 &= \tan^{-1}(2) - \tan^{-1}(1)\\ &\ \vdots\\ T_n &= \tan^{-1}(n+1) - \tan^{-1}(n) \end{align}$$ since we are adding to infinity, so all terms must cancel. whatever term we can think of will be cancelled by successive term.

so $$\sum_{i=0}^n (\tan^{-1}{(i+1)} - \tan^{-1}(i)) = - \tan^{-1}0 = 0$$

BUT

if we add all above terms, you can notice that terms start cancelling each other. so, $$\sum_{i=0}^n (\tan^{-1}{(i+1)} - \tan^{-1}(i)) = \tan^{-1}(n+1) - \tan^{-1}0$$

as $n$ tends to infinity $$\begin{align} \sum_{i=0}^\infty (\tan^{-1}{(i+1)} - \tan^{-1}(i)) &= \pi/2 - 0 \\ &= \pi/2 \end{align}$$

which sum is correct and why?

Answer 1

Sum will be $\pi/2$.

If you add, $$\lim_{n\to \infty}T_0 + T_1 + T_2 + ... T_n$$

you will get: $$\lim_{n\to \infty}\underbrace{ \tan^{-1}(1) - \tan^{-1}(0)}_{ T_0}+ ... + \underbrace{\tan^{-1}(n) - \tan^{-1}(n-1)}_{ T_{n-1}} +\underbrace{ \tan^{-1}(n + 1) - \tan^{-1}(n)}_{ T_n}$$ $$\lim_{n\to \infty}\tan^{-1}(n+1) - \tan^{-1}(0) = \pi/2$$

Answer 2

This is a telescoping series. The cancellation of terms you noticed gives $$ \sum_{k=0}^n\left(\tan^{-1}(k+1)-\tan^{-1}(k)\right)=\tan^{-1}(n+1)-\tan^{-1}(0) $$ Then note that the limit at $\infty$ is not $0$: $$ \begin{align} \sum_{k=0}^\infty\left(\tan^{-1}(k+1)-\tan^{-1}(k)\right) &=\lim_{n\to\infty}\sum_{k=0}^n\left(\tan^{-1}(k+1)-\tan^{-1}(k)\right)\\ &=\lim_{n\to\infty}\left(\tan^{-1}(n+1)-\tan^{-1}(0)\right)\\ &=\frac\pi2 \end{align} $$ You can only drop the tail terms when the limit at $\infty$ is $0$.