Arc Co Tangent Sum

Algebra Level 4

Define a sum S = ∑ i = 1 ∞ cot ⁡ − 1 2 i 2 S=\sum_{i=1}^{\infty}\cot^{-1}2i^2

Find ⌊ 1000 S ⌋ \lfloor 1000S\rfloor

Details and Assumptions

All measurements are in radians.


The answer is 785.

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Daniel Liu by Daniel Liu , 21, USA

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2 solutions

Daniel Liu
Jun 3, 2014

Here is a solution, little to no motivation included. This is because I was just playing around with telescoping series and found this sum; I have no idea how one would re-engineer this solution from scratch.

We want to create a telescoping series because the infinite sum reminds us of it. Note that something like tan ⁡ − 1 α − tan ⁡ − 1 β \tan^{-1}\alpha - \tan^{-1}\beta will telescope. We find that tan ⁡ ( tan ⁡ − 1 α − tan ⁡ − 1 β ) = α − β 1 + α β \tan(\tan^{-1}\alpha - \tan^{-1}\beta) = \dfrac{\alpha-\beta}{1+\alpha\beta} or cot ⁡ ( tan ⁡ − 1 α − tan ⁡ − 1 β ) = 1 + α β α − β \cot(\tan^{-1}\alpha - \tan^{-1}\beta) = \dfrac{1+\alpha\beta}{\alpha-\beta} or tan ⁡ − 1 α − tan ⁡ − 1 β = cot ⁡ − 1 ( 1 + α β α − β ) \tan^{-1}\alpha - \tan^{-1}\beta=\cot^{-1}\left(\dfrac{1+\alpha\beta}{\alpha-\beta}\right)

Now if we replace α = 2 i + 1 \alpha=2i+1 and β = 2 i − 1 \beta=2i-1 then we obtain tan ⁡ − 1 ( 2 i + 1 ) − tan ⁡ − 1 ( 2 i − 1 ) = cot ⁡ − 1 ( 1 + 4 i 2 − 1 2 ) = cot ⁡ − 1 ( 4 i 2 2 ) = cot ⁡ − 1 2 i 2 \begin{aligned}\tan^{-1}(2i+1)-\tan^{-1}(2i-1)&=\cot^{-1}\left(\dfrac{1+4i^2-1}{2}\right)\\ &= \cot^{-1}\left(\dfrac{4i^2}{2}\right)\\ &= \cot^{-1}2i^2\end{aligned}

which is the thing we are summing.

Thus S = ∑ i = 1 ∞ cot ⁡ − 1 2 i 2 = tan ⁡ − 1 3 − tan ⁡ − 1 1 + tan ⁡ − 1 5 − tan ⁡ − 1 3 + ⋯ → tan ⁡ − 1 ∞ − tan ⁡ − 1 1 = π 2 − π 4 = π 4 \begin{aligned}S&=\sum_{i=1}^{\infty}\cot^{-1}2i^2\\ &= \tan^{-1} 3-\tan^{-1}1+\tan^{-1}5-\tan^{-1}3+\cdots \\ &\to \tan^{-1}\infty -\tan^{-1}1\\ &= \dfrac{\pi}{2}-\dfrac{\pi}{4}\\ &= \dfrac{\pi}{4}\end{aligned}

where the arrow part means the "value" that S S approaches as the top of the summation approaches infinity. (sorry too lazy to use more proper notation; sorry about the ∞ \infty part too)

Thus our answer is ⌊ 1000 S ⌋ = ⌊ 250 π ⌋ = 785 \lfloor 1000S\rfloor = \lfloor 250\pi \rfloor = \boxed{785}

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That's often the case with telescoping series, where it's not obvious how / why you should create those sequences r n = t n − t n − 1 r_n = t_n - t_{n-1} .

Let me post another one where it's not obvious what the t n t_n is.

Calvin Lin Staff - 7 years ago

Converting 2 i 2 2i^{2} into 4 i 2 2 \dfrac{4i^{2}}{2} is absolutely necessary. Or else, one cannot convert into a telescoping series.

Avineil Jain - 7 years ago

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Can you please explain to me how you solved the problem? Thanks.

Daniel Liu - 7 years ago

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I just solved the reverse way you did.

∑ i = 0 ∞ c o t − 1 2 i 2 \displaystyle\sum_{i=0}^∞ cot^{-1} 2i^{2}

∑ i = 0 ∞ c o t − 1 4 i 2 2 \displaystyle\sum_{i=0}^∞ cot^{-1}\dfrac{4i^{2}}{2}

∑ i = 0 ∞ t a n − 1 2 4 i 2 \displaystyle\sum_{i=0}^∞ tan^{-1}\dfrac{2}{4i^{2}}

∑ i = 0 ∞ t a n − 1 ( 2 i + 1 ) − ( 2 i − 1 ) 1 + ( 2 i + 1 ) ( 2 i − 1 ) \displaystyle\sum_{i=0}^∞ tan^{-1}\dfrac{(2i + 1) - (2i - 1)}{1 + (2i + 1)(2i - 1)}

∑ i = 0 ∞ t a n − 1 ( 2 i + 1 ) − t a n − 1 ( 2 i − 1 ) \displaystyle\sum_{i=0}^∞ tan^{-1}(2i + 1) - tan^{-1} (2i -1)

Note that conversion of 2 i 2 2i^{2} into 4 i 2 2 \dfrac{4i^{2}}{2} brought on the correct solution. Or else, can the series be made only with 2 i 2 2i^{2} ???

Avineil Jain - 7 years ago
Ariel Gershon
Sep 11, 2014

I say that the n t h n^{th} partial sum equals a r c t a n ( n n + 1 ) arctan \left(\frac{n}{n+1} \right) . The base case n = 1 n = 1 is true since a r c c o t ( 2 ) = a r c t a n ( 1 2 ) arccot(2) = arctan \left( \frac{1}{2} \right) .

Now suppose the k t h k^{th} partial sum equals a r c t a n ( k k + 1 ) arctan \left(\frac{k}{k+1} \right) and consider the ( k + 1 ) t h (k+1)^{th} partial sum:

∑ i = 1 k + 1 a r c c o t ( 2 i 2 ) = a r c t a n ( k k + 1 ) + a r c c o t ( 2 ( k + 1 ) 2 ) \sum_{i = 1}^{k+1} arccot(2i^2) = arctan \left(\frac{k}{k+1} \right) + arccot(2(k+1)^2) = a r c t a n ( k k + 1 ) + a r c t a n ( 1 2 ( k + 1 ) 2 ) = arctan \left(\frac{k}{k+1} \right) + arctan \left(\frac{1}{2(k+1)^2}\right)

Now let α = a r c t a n ( k k + 1 ) \alpha = arctan \left(\frac{k}{k+1} \right) and β = a r c t a n ( 1 2 ( k + 1 ) 2 ) \beta = arctan \left(\frac{1}{2(k+1)^2}\right) . Using a well-known trigonometric identity, we see that: t a n ( α + β ) = t a n ( α ) + t a n ( β ) 1 − t a n ( α ) t a n ( β ) = k k + 1 + 1 2 ( k + 1 ) 2 1 − k k + 1 ∗ 1 2 ( k + 1 ) 2 = 2 k ( k + 1 ) 2 + ( k + 1 ) 2 ( k + 1 ) 3 − k tan(\alpha + \beta) = \frac{tan(\alpha) + tan(\beta)}{1 - tan(\alpha)tan(\beta)} = \frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1 - \frac{k}{k+1}*\frac{1}{2(k+1)^2}} = \frac{2k(k+1)^2 + (k+1)}{2(k+1)^3 - k} = 2 k 3 + 4 k 2 + 3 k + 1 2 k 3 + 6 k 2 + 5 k + 2 = ( k + 1 ) ( 2 k 2 + 2 k + 1 ) ( k + 2 ) ( 2 k 2 + 2 k + 1 ) = k + 1 k + 2 =\frac{2k^3+4k^2+3k+1}{2k^3+6k^2+5k+2} = \frac{(k+1)(2k^2+2k+1)}{(k+2)(2k^2+2k+1)} = \frac{k+1}{k+2}

Therefore t a n ( α + β ) = k + 1 k + 2 tan(\alpha + \beta) = \frac{k+1}{k+2} . Since 0 < k k + 1 < 1 0 < \frac{k}{k+1} < 1 and 0 < 1 2 ( k + 1 ) 2 < 1 0 < \frac{1}{2(k+1)^2} < 1 , this means that 0 < α < π 4 0 < \alpha < \frac{\pi}{4} and 0 < β < π 4 0 < \beta < \frac{\pi}{4} . Therefore 0 < α + β < π 2 0 < \alpha + \beta < \frac{\pi}{2} . Therefore α + β = a r c t a n ( k + 1 k + 2 ) \alpha + \beta = arctan \left(\frac{k+1}{k+2} \right) . Hence the ( k + 1 ) t h (k+1)^{th} partial sum is a r c t a n ( k + 1 k + 2 ) arctan \left(\frac{k+1}{k+2} \right) , so the induction is complete.

Therefore, ∑ i = 1 ∞ a r c c o t ( 2 i 2 ) = lim ⁡ n → ∞ a r c t a n ( n n + 1 ) = a r c t a n ( 1 ) = π 4 \sum_{i = 1}^{\infty} arccot(2i^2) = \lim_{n \to \infty} arctan \left(\frac{n}{n+1} \right) = arctan(1) = \boxed{\frac{\pi}{4}}

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