sequence & series # logarithmic series question

Question

The series 2[$\frac{1}{3x+1}$ + $\frac{1}{3(3x+1)^3}$ + $\frac{1}{5(3x+1)^5}$ + ...] is equal to

Answer 1

We know that $$\ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\ldots\right)$$

Putting $x=\frac{1}{3x+1}$; $$\ln\left(1+\frac{2}{3x}\right)=2\left(\frac{1}{3x+1} + \frac{1}{3(3x+1)^3} + \frac{1}{5(3x+1)^5} + \ldots\right)$$

Answer 2

Use What is the correct radius of convergence for $\ln(1+x)$?

for $-1<y<1$

$$\ln(1+y)-\ln(1-y)=2\left(\sum_{r=0}^\infty\dfrac{y^{2r+1}}{2r+1}\right)$$

Answer 3

The Taylor series for $\tan^{-1} x$ is:

$$x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots + (-1)^{n} \frac{x^{2n+1}}{2n+1}$$

Now use the fact that $\tanh^{-1} x = \frac{1}{i} \tan^{-1} (ix)$. (Wolfram MathWorld)