# How do you find the maclaurin series expansion of #ln((1+x)/(1-x))#?

To accomplish this, we write

We would need to compute some derivatives, as this infinite sum implies. I will do this for three terms, hopefully enough to identify a pattern.

Due to logarithm properties, we have

Calculating derivatives:

Observing that some terms simply cancel each other out, we are left with

When we reduce the denominators of each term (the factorials), we obtain

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To find the Maclaurin series expansion of (\ln\left(\frac{1+x}{1-x}\right)), we can start by using the properties of logarithms to rewrite the expression as (\ln(1+x) - \ln(1-x)). Then, we can use the Maclaurin series expansions of (\ln(1+x)) and (\ln(1-x)), which are:

[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots ] [ \ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots ]

Substitute these series expansions into (\ln\left(\frac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x)) and simplify to get the Maclaurin series expansion.

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