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Question 11 Find the sum: \sum_{n=0}^{\infty} \frac{(-1)^n}{({\sqrt{3}})^{2n+1}(2n+1)} I wrote a solution. I did not write a solution. tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}

          Question 11
Find the sum: \sum_{n=0}^{\infty} \frac{(-1)^n}{({\sqrt{3}})^{2n+1}(2n+1)}
I wrote a solution.
I did not write a solution.
tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}

Question 11
Find the sum: ∑n=0^∞ ((-1)^n)/((√(3))^2n+1(2n+1))
I wrote a solution.
I did not write a solution.
tan^-1((1)/(√(3))) = (π)/(6)

Added by Richard S.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Wendi Zhao

06/07/2021

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Step 1: The given series is an infinite geometric series. Show more…

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Question 11 Find the sum: ∑n=0∞ (-1)^n / (√3)^(2n+1)(2n+1) I wrote a solution. I did not write a solution. tan^-1(1/√3) = π/6

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Question 11 Find the sum: \sum_{n=0}^{\infty} \frac{(-1)^n}{({\sqrt{3}})^{2n+1}(2n+1)} I wrote a solution. I did not write a solution. tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}

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