Question

Problem Use the formula: $\tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right)$ $-1 < x < 1$ and the Maclaurin series for $\ln(1+x)$ to show that $\tanh^{-1} x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}$

          Problem Use the formula:
$\tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right)$  $-1 < x < 1$
and the Maclaurin series for $\ln(1+x)$ to show that
$\tanh^{-1} x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}$

Problem Use the formula:
tanh^-1 x = (1)/(2)ln( (1+x)/(1-x)) -1 < x < 1
and the Maclaurin series for ln(1+x) to show that
tanh^-1 x = ∑n=0^∞(x^2n+1)/(2n+1)

Added by Larry M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Raj Bala

01/12/2022

Step 1

Step 1: Start with the given formula ln(1+x)tanh^(-1)x=\sum_(n=0)^(∞) (x^(2n+1))/(2n+1)tanh^(-1)x=(1)/(2)ln((1+x)/(1-x)),-1 and the Maclaurin series for ln(1+x). Show more…

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Problem Use the formula: ln(1+x)tanh^(-1)x=sum_(n=0)^(infty ) (x^(2n+1))/(2n+1)tanh^(-1)x=(1)/(2)ln((1+x)/(1-x)),-1 and the Maclaurin series for ln(1+x) to show that tanh^(-1)x=sum_(n=0)^(infty ) (x^(2n+1))/(2n+1) Problem Use the formula: 1+ tanh-1=in and the Maclaurin series for ln(1+ x) to show that -1x1 x2n+1 2n+1 tanh-1