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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 $\infty$.

Thank you for your help.

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  • $\begingroup$ $$\sum\limits_{n=1}^k[\arctan(2n+1)-\arctan(2n-1)]=$$ $$(\arctan(3)-\arctan(1))+(\arctan(5)-\arctan(3))+(\arctan(7)-\arctan(5))+\cdots+(\arctan(2k-1)-\arctan(2k-3))+(\arctan(2k+1)-\arctan(2k-1)) =$$ $$=\arctan(2k+1)-\arctan(1)$$ $\endgroup$ Jan 25 at 10:49
  • $\begingroup$ @TitoEliatron right. But our series is infinite. $\endgroup$ Jan 25 at 10:50
  • $\begingroup$ 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 $\arctan(+\infty)-\arctan(1)=\pi/2-\pi/4=\pi/4$. $\endgroup$ Jan 25 at 10:50
  • $\begingroup$ @A-LevelStudent Why does finiteness matter? How is it any different from taking the limit $k \to \infty$? $\endgroup$
    – Ankit Saha
    Jan 25 at 10:52
  • $\begingroup$ @AnkitSaha with the infinite series group the terms differently: $$-\arctan 1+(\arctan 3-\arctan3)+(\arctan5-\arctan5)+\cdots+(\arctan(k)-\arctan(k))+\cdots$$- $\endgroup$ Jan 25 at 10:57
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A few intermediate steps might be helpful.

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*}

Comment:

  • In (1) we apply telescoping.

  • In (2) we use $\lim_{N\to\infty}\arctan(N)=\frac{\pi}{2}$ and $\lim_{N\to\infty} a=a$.

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