Question #7f415

Question

1129 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-14T23:36:58

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

Let #g(t)=ln(1+t) Rightarrow g(0)=0#

By differentiating w.r.t. #t#,

#g'(t)=1/(1+t)=1/(1-(-t))#

By the sum of geometric series #a/(1-r)=sum_(n=0)^(infty)ar^n#,

#g'(t)=sum_(n=0)^(infty)1cdot(-t)^n=sum_(n=0)^(infty)(-1)^n t^n#

By integrating w.r.t. #t#,

#g(t)=sum_(n=0)^(infty)(-1)^n t^(n+1)/(n+1)+C#,

where #C=0# since #g(0)=0#

So, we have

#ln(1+t)=sum_(n=0)^(infty)(-1)^n t^(n+1)/(n+1)#

By setting #t=x-1#,

#ln(x)=sum_(n=0)^(infty)(-1)^n (x-1)^(n+1)/(n+1)#

I hope that this was clear.