(1) #f(x) = ln(x) implies f(1)=0#
#f^'(x) = 1/x implies f^'(1) = 1#
#f''(x) = -1/x^2 implies f''(1) = -1#
#f'''(x) = ((-1)(-2))/x^3 implies f'''(1) = 2#
#f^((4))(x) = ((-1)(-2)(-3))/x^4 implies f^((4))(1) = -6#
The general pattern should be clear from this:
#f^((n))(x) = ((-1)(-2)...(-n+1))/x^n implies#
# f^((n))(1) =(-1)^(n-1)(n-1)! #
(2) So, the Taylor series for #ln(x)# centered at #x=1# is given by
#ln (x) = sum_{n=0}^oo f^((n))(1)/(n!) (x-1)^n#
#qquadqquad = sum_{n=0}^oo (-1)^(n-1)((n-1)!)/(n!) (x-1)^n#
#qquad qquad = sum_{n=0}^oo (-1)^{n-1} (x-1)^n/n#
(3) The #n#-th term in the Taylor series above is given by
#t_n = (-1)^(n-1)(x-1)^n/n#
and thus
#t_{n+1}/t_n = -n/(n+1)(x-1)#
For the series to converge, we must have
#lim_{n to oo}|t_{n+1}/t_n|<1 implies#
#lim_{n to oo}|-n/(n+1)(x-1)|<1 implies#
#color(red)(|x-1|<1)#
which is the domain of convergence.
(4)
#ln(4/3) = sum_{n=0}^oo (-1)^{n-1} (4/3-1)^n/n#
#qquad qquad = sum_{n=0}^oo (-1)^{n-1} 1/(n3^n)#
#ln(2/3) = sum_{n=0}^oo (-1)^{n-1} (2/3-1)^n/n#
#qquad qquad = sum_{n=0}^oo (-1)^{n-1} (-1)^n/(n3^n)#
Hence
#ln (2) = ln(4/3)-ln(2/3) #
#qquad = sum_{n=0}^oo (-1)^{n-1} (1-(-1)^n)/(n3^n)#
In this sum, all the even terms cancel, so that we get
#ln (2) = sum_{n=0}^oo (-1)^{2n+1-1} 2/((2n+1)3^(2n+1))#
and hence
#ln(sqrt2) = 1/2 ln(2) = sum_{n=0}^oo 1/((2n+1)3^(2n+1))#
