Taylor y arc tan = tan ⁡ − 1 \tan^{-1}

Calculus Level 4

∑ n = 1 ∞ ( − 1 ) n 4 n ( 2 n + 1 ) \displaystyle \sum_{n = 1}^\infty \frac{(-1)^n}{4^n(2n + 1)} The sum above can be written as S = A arctan ⁡ ( 1 A ) − B S=A\arctan \left(\dfrac{1}{A}\right) - B where A A and B B are coprime positive integers. Find 10 A + B 10A + B .


The answer is 21.

Guillermo Templado by Guillermo Templado , 46, Spain

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1 solution

Otto Bretscher
Apr 29, 2016

Great problem for a calculus exam!

We have the Taylor series arctan ⁡ ( x ) = x + ∑ n = 1 ∞ ( − 1 ) n 2 n + 1 x 2 n + 1 \arctan(x)=x+\sum_{n=1}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1} for ∣ x ∣ < 1 |x|<1 . Evaluating at x = 1 2 x=\frac{1}{2} gives arctan ⁡ ( 1 2 ) = 1 2 + ∑ n = 1 ∞ ( − 1 ) n 2 × 4 n ( 2 n + 1 ) \arctan(\frac{1}{2})=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{(-1)^n}{2\times4^n(2n+1)} so S = 2 arctan ⁡ ( 1 2 ) − 1 S=2\arctan(\frac{1}{2})-1 and the answer is 21 \boxed{21}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Perfect (+1). Same aprroach than me...

Guillermo Templado - 5 years, 1 month ago

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