Infinite Arccotangents Equals One Arctangent

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

$\begin{aligned}\mathfrak{T}=&\displaystyle\sum_{r=1}^n\cot^{-1}\left(r^2+\dfrac 34\right)\\=&\displaystyle\sum_{r=1}^n\tan^{-1}\left( \dfrac{1}{r^2+\dfrac 34}\right)\\=&\displaystyle\sum_{r=1}^n\tan^{-1} \left(\dfrac{1}{ 1+\left(r^2-\dfrac 14\right)}\right)\\=&\displaystyle\sum_{r=1}^n\tan^{-1}\left( \dfrac{\left(r+\frac 12\right)-\left(r-\frac 12\right)}{1+\left(r+\frac 12\right)\left(r-\frac 12\right)}\right)\\=&\displaystyle\sum_{r=1}^n\left[\tan^{-1}\left(r+\frac 12\right)-\tan^{-1}\left(r-\frac 12\right)\right]\end{aligned}$

$\large\mathbf{(A~Telescopic~Series)}$

$\large=\tan^{-1}\left(n+\frac 12\right)-\tan^{-1}\frac 12$

$\color{#0C6AC7}{\large\therefore\displaystyle\lim_{n\to\infty}\mathfrak{T}=\dfrac{\pi}{2}-\tan^{-1}\frac 12=\tan^{-1}2}$

$\cot^{-1}\left(1^2+\dfrac{3}{4}\right)+\cot^{-1}\left(2^2+\dfrac{3}{4}\right)+\cot^{-1}\left(3^2+\dfrac{3}{4}\right)+\cot^{-1}\left(4^2+\dfrac{3}{4}\right)+\ldots\\ =\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \cot^{-1}\left(\dfrac{4k^2+3}{4}\right)\\ =\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \color{#3D99F6}{\tan^{-1}\left(\dfrac{4}{4k^2+3}\right)}\\ =\displaystyle\lim_{n \to \infty} \color{#D61F06}{\tan^{-1}\left(\dfrac{4n}{5+2n}\right)}\\ =\displaystyle\lim_{n \to \infty} \tan^{-1}\left(\dfrac{4}{\frac{5}{n}+2}\right)\\ =\tan^{-1} \left(\dfrac{4}{0+2}\right)\\ =\boxed{\tan^{-1} 2}$

The proof for both highlighted expressions are found below

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Proof 1:

$\color{#3D99F6}{\cot^{-1} a = \tan^{-1}\left(\dfrac{1}{a}\right)}$

Let $\cot^{-1} a= b$

$a=\cot b\\ a=\dfrac{1}{\tan b}\\ \tan b= \dfrac{1}{a}\\ b = \tan^{-1}\left(\dfrac{1}{a}\right)\\ \color{#3D99F6}{\cot^{-1} a =\tan^{-1}\left(\dfrac{1}{a}\right)}$

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Proof 2:

$\color{#D61F06}{\sum_{k=1}^n \tan^{-1}\left(\dfrac{4}{4k^2+3}\right)=\tan^{-1}\left(\dfrac{4n}{5+2n}\right)}$

First of all, we need to show that:

$\color{#20A900}{\tan^{-1} a+ \tan^{-1} b = \tan^{-1}\left(\dfrac{a+b}{1-ab}\right)}$

Let $\tan^{-1} a = x,\;\tan^{-1} b = y$

$\tan^{-1} a+ \tan^{-1} b = x+y\\ \tan(x+y) = \tan\left(\tan^{-1} a+ \tan^{-1} b\right)\\ \tan(x+y)=\dfrac{\tan\left(\tan^{-1} a\right)+\tan\left( \tan^{-1} b\right)}{1-\tan\left(\tan^{-1} a\right)\tan\left( \tan^{-1} b\right)}\\ \tan(x+y)=\dfrac{a+b}{1-ab}\\ x+y=\tan^{-1}\left(\dfrac{a+b}{1-ab}\right)\\ \color{#20A900}{\tan^{-1} a+ \tan^{-1} b = \tan^{-1}\left(\dfrac{a+b}{1-ab}\right)}$

To see this pattern, let us try to get the sum for $n=1,2,3,4$

$n=1 \implies\\ \displaystyle\sum_{k=1}^1 \tan^{-1}\left(\dfrac{4}{4k^2+3}\right)\\ =\tan^{-1}\left(\dfrac{4}{7}\right)\\ n=2 \implies\\ \displaystyle\sum_{k=1}^2 \tan^{-1}\left(\dfrac{4}{4k^2+3}\right)\\ =\tan^{-1}\left(\dfrac{4}{7}\right)+\tan^{-1}\left(\dfrac{4}{19}\right)\\ =\tan^{-1}\left(\dfrac{\frac{4}{7}+\frac{4}{19}}{1-\frac{4}{7}\times\frac{4}{19}}\right)\\ =\tan^{-1}\left(\dfrac{8}{9}\right)\\ n=3 \implies\\ \displaystyle\sum_{k=1}^3 \tan^{-1}\left(\dfrac{4}{4k^2+3}\right)\\ =\color{#69047E}{\tan^{-1}\left(\dfrac{4}{7}\right)+\tan^{-1}\left(\dfrac{4}{19}\right)}+\tan^{-1}\left(\dfrac{4}{39}\right)\\ =\tan^{-1}\left(\dfrac{8}{9}\right)+\tan^{-1}\left(\dfrac{4}{39}\right)\\ =\tan^{-1}\left(\dfrac{\frac{8}{9}+\frac{4}{39}}{1-\frac{8}{9}\times\frac{4}{39}}\right)\\ =\tan^{-1}\left(\dfrac{12}{11}\right)\\ n=4 \implies\\ \displaystyle\sum_{k=1}^4 \tan^{-1}\left(\dfrac{4}{4k^2+3}\right)\\ =\color{#69047E}{\tan^{-1}\left(\dfrac{4}{7}\right)+\tan^{-1}\left(\dfrac{4}{19}\right)+\tan^{-1}\left(\dfrac{4}{39}\right)}+\tan^{-1}\left(\dfrac{4}{67}\right)\\ =\tan^{-1}\left(\dfrac{12}{11}\right)+\tan^{-1}\left(\dfrac{4}{67}\right)\\ =\tan^{-1}\left(\dfrac{\frac{12}{11}+\frac{4}{67}}{1-\frac{12}{11}\times\frac{4}{67}}\right)\\ =\tan^{-1}\left(\dfrac{16}{13}\right)$

Notice that for the first $4$ values of $n$ , the sum gives you $\tan^{-1}\left(\dfrac{4n}{5+2n}\right)$

This relationship can be further proved with induction.

Proof by induction

Let $P_n$ be the proposition that $\displaystyle\color{#D61F06}{\sum_{k=1}^n \tan^{-1}\left(\dfrac{4}{4k^2+3}\right)=\tan^{-1}\left(\dfrac{4n}{5+2n}\right)}$ for all positive integers $n$ .

Base case: Let $n=1$

LHS $=\tan^{-1}\left(\dfrac{4}{4(1)^2+3}\right)=\tan^{-1}\left(\dfrac{4}{7}\right)$

RHS $=\tan^{-1}\left(\dfrac{4(1)}{5+2(1)}\right)=\tan^{-1}\left(\dfrac{4}{7}\right)$

LHS = RHS. Base case holds true.

Induction Hypothesis: Suppose $P_m$ is true for some integer $m$ . Consider $P_{m+1}$

LHS $P_{m+1}$

$=\tan^{-1}\left(\dfrac{4m}{5+2m}\right)+\tan^{-1}\left(\dfrac{4}{4(m+1)^2+3}\right)\\ =\tan^{-1}\left(\dfrac{\frac{4m}{5+2m}+\frac{4}{4m^2+8m+7}}{1-\left(\frac{4m}{5+2m}\right)\left(\frac{4}{4m^2+8m+7}\right)}\right)\\ =\tan^{-1}\left(\dfrac{\left(\frac{4m(4m^2+8m+7)+4(5+2m)}{(5+2m)(4m^2+8m+7)}\right)}{\left(\frac{(5+2m)(4m^2+8m+7)-16m}{(5+2m)(4m^2+8m+7)}\right)}\right)\\ =\tan^{-1}\left(\dfrac{16m^3+32m^2+36m+20}{8m^3+36m^2+38m+35}\right)\\ =\tan^{-1}\left(\dfrac{4(4m^3+8m^2+9m+5)}{8m^3+28m^2+8m^2+28m+10m+35}\right)\\ =\tan^{-1}\left(\dfrac{4(4m^3+4m^2+4m^2+4m+5m+5)}{4m^2(2m+7)+4m(2m+7)+5(2m+7)}\right)\\ =\tan^{-1}\left(\dfrac{4\left(4m^2(m+1)+4m(m+1)+5(m+1)\right)}{(2m+7)(4m^2+4m+5)}\right)\\ =\tan^{-1}\left(\dfrac{4(m+1)(4m^2+4m+5)}{(5+2m+2)(4m^2+4m+5)}\right)\\ =\tan^{-1}\left(\dfrac{4(m+1)}{5+2(m+1)}\right)$

$=$ RHS $P_{m+1}$

Conclusion: By Mathematical Induction, since $P_1$ is true and $P_m$ is true implies $P_{m+1}$ is true, thus $P_n$ is true for all positive integers $n$

Phew! That was a long proof