How do you derive the maclaurin series for $f(x)=ln(1+x)$ ?

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Answer 1 · 2017-04-17T23:52:58

$sum_(n=1)^(infty)(-1)^(n-1)x^n/n$

$f(x)=ln(1+x)$

By differentiating w.r.t. $x$ ,

$f'(x)=1/(1+x)=1/(1-(-x))$

By viewing it as the sum of a geometric series,

$f'(x)=sum_(n=0)^(infty)1cdot(-x)^n=sum_(n=0)^(infty)(-1)^nx^n$

By integrating w.r.t $x$ term by term,

$f(x)=sum_(n=0)^(infty)(-1)^n(x^(n+1))/(n+1)+C$

Since $f(0)=0$ gives $C=0$ ,

$f(x)=sum_(n=0)^(infty)(-1)^n(x^(n+1))/(n+1)$

By shifting the indices by $1$ ,

$f(x)=sum_(n=1)^(infty)(-1)^(n-1)x^n/n$

I hope that this was clear.

How do you derive the maclaurin series for f(x)=ln(1+x)f(x)=ln(1+x)?