Sum With Hidden arctan(x)

Calculus Level 4

If $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)} = \frac{\pi}{A}-\frac{\ln|B|}{B}$ , find the minimum value of $A+B$ .

The answer is 6.

2 solutions

Chew-Seong Cheong
Mar 26, 2019

$\begin{aligned} S & = \sum_{n=0}^\infty \frac {(-1)^n}{(2n+1)(2n+2)} \\ & = \sum_{n=0}^\infty (-1)^n \left(\frac 1{2n+1} - \frac 1{2n+2} \right) \\ & = \sum_{n=0}^\infty \frac {(-1)^n}{2n+1} - \frac 12 \sum_{n=0}^\infty \frac {(-1)^n}{n+1} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \tan^{-1} 1 - \frac {\ln 2}2 \\ & = \frac \pi 4 - \frac {\ln 2}2 \end{aligned}$

Therefore, $A+B = 4+2 = \boxed 6$ .

Reference: Maclaurin series

Thank you. Your solution is much more elegant and simple than mine. I ended up using the integral of the Taylor Series of arctan(x) to get the solution.

Solden Stoll - 2 years, 2 months ago

You are welcome. I will learn a lot here with Brilliant.org.

Chew-Seong Cheong - 2 years, 2 months ago

I am a moderator. I have changed the wording of your problem because $\dfrac {\ln 4}4 = \dfrac {\ln 2}2$ and hence $B=4$ is also a solution.

Chew-Seong Cheong - 2 years, 2 months ago

Solden Stoll
Mar 25, 2019

Hint: Remember $\arctan(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$

$B=4$ is also a solution.

Chew-Seong Cheong - 2 years, 2 months ago

Sum With Hidden arctan(x)

The answer is 6.

2 solutions

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