Best solutions to - "How to sum up this inverse function limn→∞∑r=0ntan−1(2r1+2r4)"

derlingasmh

Answered question

2022-01-23

How to sum up this inverse function

lim_{n \to \infty} \sum_{r = 0}^{n} \tan^{- 1} (\frac{2 r}{1 + 2 r^{4}})

Emilie Booker

Beginner2022-01-24Added 14 answers

I'll leave it to you to prove by induction that the partial sum is

\arctan (1 - \frac{1}{n^{2} + n + 1})

, so the limit is

\frac{π}{4}

. One approach to obtaining this partial sum is that of hint(

(x, y) = (f (r - 1), f (r)) where f (r) = 2 r^{2} + 2 r + 1

)

{[\arctan (2 r^{2} + 2 r + 1)]}_{- 1}^{n} = \arctan \frac{n^{2} + n}{n^{2} + n + 1}

Edit: just to spell it out, the definition

f (r) = \arctan (2 r^{2} + 2 r + 1)

gives

\sum_{r = 0}^{n} \arctan \frac{2 r}{1 + 2 r^{4}} = \sum_{r = 0}^{n} (f (r) - f (r - 1)) = f (n) - f (- 1)

where the second = can be proven by induction; this is the basis of any proof with telescoping series. In particular

lim_{n \to \infty} (f (n) - f (- 1)) = \arctan \infty - \arctan 1 = \frac{π}{2} - \frac{π}{4} = \frac{π}{2}

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