# Telescoping series: $\sum\limits_{n=1}^{∞}[\tan^{-1}(2n+1)-\tan^{-1}(2n-1)]$

In this question that was asked today the OP wrote that \begin{align}\sum\limits_{n=1}^{∞}[\arctan(2n+1)-\arctan(2n-1)]&=\arctan\infty-\arctan 1\\ &=\frac{\pi}{2}-\frac{\pi}{4}\\&=\frac{\pi}{4} \end{align}

I don't really understand why there is a $$\arctan\infty$$ term. Surely every single term other than $$-\arctan 1$$ has been cancelled out? Now I know that I am wrong, as obviously the value of the summation cannot be negative, but I'm not sure where I am wrong.

I have thought of considering the similar finite series, $$\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=\arctan(2k+1)-\arctan1$$ and letting $$k\to\infty$$, so that $$\lim_{k\to\infty}\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=\arctan\infty-\arctan1$$ as required, but I still can't see why all the terms other than $$-\arctan1$$ don't cancel out, as the upper limit actually is $$\infty$$.

• $$\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=$$ $$(\arctan(3)-\arctan(1))+(\arctan(5)-\arctan(3))+(\arctan(7)-\arctan(5))+\cdots+(\arctan(2k-1)-\arctan(2k-3))+(\arctan(2k+1)-\arctan(2k-1)) =$$ $$=\arctan(2k+1)-\arctan(1)$$ Jan 25 at 10:49
• @TitoEliatron right. But our series is infinite. Jan 25 at 10:50
• This is the $k$-th PARTIAL SUM. The series converges iff the sequence of partial sums converges. And in this particular case, the sequence $S_k$ of partial sums converges to $\arctan(+\infty)-\arctan(1)=\pi/2-\pi/4=\pi/4$. Jan 25 at 10:50
• @A-LevelStudent Why does finiteness matter? How is it any different from taking the limit $k \to \infty$? Jan 25 at 10:52
• @AnkitSaha with the infinite series group the terms differently: $$-\arctan 1+(\arctan 3-\arctan3)+(\arctan5-\arctan5)+\cdots+(\arctan(k)-\arctan(k))+\cdots$$- Jan 25 at 10:57

We obtain \begin{align*} \color{blue}{\sum_{n=1}^{\infty}}&\color{blue}{\left[\arctan(2n+1)-\arctan(2n-1)\right]}\\ &=\lim_{N\to\infty}{\sum_{n=1}^{N}\left[\arctan(2n+1)-\arctan(2n-1)\right]}\\ &=\lim_{N\to\infty}\left[\arctan(2N+1)-\arctan(1)\right]\tag{1}\\ &=\lim_{N\to\infty}\arctan(2N+1)-\lim_{N\to\infty}\arctan(1)\tag{2}\\ &=\frac{\pi}{2}-\frac{\pi}{4}\\ &\,\,\color{blue}{=\frac{\pi}{4}}\\ \end{align*}
• In (2) we use $$\lim_{N\to\infty}\arctan(N)=\frac{\pi}{2}$$ and $$\lim_{N\to\infty} a=a$$.