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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.

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  • $\begingroup$ How do you get $\tan^{-1}(-2)$ term? $\endgroup$
    – Bob Dobbs
    Jun 9, 2023 at 15:56
  • $\begingroup$ @BobDobbs it was a typo, corrected it $\endgroup$ Jun 9, 2023 at 17:37
  • $\begingroup$ WolframAlpha says the answer is $\pi/2$ so that means we must take the typo $\tan^{-1}(-1)$ as $3\pi/4$? @failedC $\endgroup$
    – Bob Dobbs
    Jun 9, 2023 at 19:07
  • $\begingroup$ $\arctan\left(\frac{3}{n^{2}+n-1}\right) \neq \arctan\left(n+2\right)-\arctan\left(n-1\right)$ when $n = 0$, but for all the other positive integers $n$, it holds. $\endgroup$ Jun 10, 2023 at 1:48
  • $\begingroup$ @BobDobbs due to a particularly stupid mistake, I've written n=0 instead of n=1, but I've understood MathFail's answer. Thanks for your interest $\endgroup$ Jun 10, 2023 at 4:26

3 Answers 3

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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

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

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  • $\begingroup$ OP says "when summing to infinity why do we have remanent", so I guess it is confused on the partial sum and series notations. $\endgroup$
    – MathFail
    Jun 9, 2023 at 16:03
  • $\begingroup$ why is the series equal to the limit of the partial sum. This is obviously not the answer but when we write down the telescoping summation, all the terms apart from $-\tan^{-1} 1 - \tan^{-1} 2$ appear to get cancelled $\endgroup$ Jun 9, 2023 at 17:44
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    $\begingroup$ Series is defined by the limit of partial sum. So the formal way is to first evaluate the partial sum to some finite terms. In your OP, the partial sum is to evaluate it from $k=0$ to $k=n$, so totally $n+1$ terms, then you do the telescope and cancel those middle terms, and left with 6 terms. Finally take the limit by letting $n\to\infty$ $\endgroup$
    – MathFail
    Jun 9, 2023 at 17:52
  • $\begingroup$ OP changed the question. $\endgroup$
    – Bob Dobbs
    Jun 10, 2023 at 5:43
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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 $$S=\frac{3\pi}{2}-\tan^{-1}2-\frac{\pi}{4}=\frac{5\pi}{4}-\tan^{-1}2.$$

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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

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

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  • $\begingroup$ OP changed the question. $\endgroup$
    – Bob Dobbs
    Jun 10, 2023 at 5:43

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