**(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))#