Remember the MacLaurin Series:
#color(blue)(ln(1+u)=sum_(n=0)^(oo)frac{(-1)^n(u)^(n+1)}{n+1})#
Change the index:
#color(blue)(=sum_(n=1)^(oo)frac{(-1)^(n+1)(u)^(n)}{n})#
Substitute #u=4x# because
#f(x)=ln(1+4x)#
#ln(1+4x)=sum_(n=1)^(oo)frac{(-1)^(n+1)(4x)^n}{n}#
#=sum_(n=1)^(oo)frac{(-1)^(n+1)(4)^n(x)^n}{n}#
To find radius of convergence, use the ratio test: #color(blue)(lim_(n->oo)frac{a_(n+1)}{a_n})#
#lim_(n->oo)|frac{(-1)^(n+2)(4)^(n+1)(x)^(n+1)}{n+1}*frac{n}{(-1)^(n+1)(4)^n(x)^n}|#
#=lim_(n->oo)|frac{(-1)^(n+2)cancel((4)^(n))(4)cancel((x)^(n))(x)}{n+1}*frac{n}{(-1)^(n+1)cancel((4)^n)cancel((x)^n)}|#
(Note that #(-1)^n# disappears because of the absolute value)
#=|4x|#
For this to converge, #|4x|<1#
#4x<1#
#x<1/4#
#4x > -1#
#x > -1/4#
Therefore, the radius of convergence (#R#) of the MacLaurin representation of #f(x)=ln(1+4x)# is equal to #1/4#
