We have an answer to "The series 2[13x+1+13(3x+1)3+15(3x+1)5+…] is equal to?"

Brock Brown

Answered question

2021-12-23

The series

2 [\frac{1}{3 x + 1} + \frac{1}{3 {(3 x + 1)}^{3}} + \frac{1}{5 {(3 x + 1)}^{5}} + \dots]

is equal to?

Answer & Explanation

Chanell Sanborn

Beginner2021-12-24Added 41 answers

We know that

\ln (\frac{1 + x}{1 - x}) = 2 (x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + \dots)

Putting

x = \frac{1}{3 x + 1}

;

\ln (1 + \frac{2}{3 x}) = 2 (\frac{1}{3 x + 1} + \frac{1}{3 {(3 x + 1)}^{3}} + \frac{1}{5 {(3 x + 1)}^{5}} + \dots)

Stuart Rountree

Beginner2021-12-25Added 29 answers

For

- 1 < y < 1

\ln (1 + y) - \ln (1 - y) = 2 (\sum_{r = 0}^{\infty} \frac{y^{2 r + 1}}{2 r + 1})

user_27qwe

Skilled2021-12-30Added 375 answers

The Taylor series for $\tan^{- 1} x$ is:
$x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - . . . + (- 1)^{n} \frac{x^{2 n + 1}}{2 n + 1}$
Now use the fact that $\tanh^{- 1} x = \frac{1}{i} \tan^{- 1} (i x)$

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