What is the interval of convergence of #sum_1^oo (2^k)/k (x-1)^k #?

1 Answer
Cesareo R.
May 25, 2016

#sum_{i=1}^{infty}(2^i(x-1)^i)/i = log_e (1/(3-2x))# for #abs(x-1)<1/2#

Explanation:

Let #f(x)=sum_{i=1}^{infty}(2^i(x-1)^i)/i#
calculating
#(df)/(dx)(x) = sum_{i=1}^{infty}(2^i(x-1)^{i-1})# simplifying
#(df)/(dx)(x) = 2sum_{j=0}^{infty}(2(x-1))^j#
following the polynomial identity
#(1-z^{n+1})/(1-z)=1+z+z^2+z^3+...+ z^n#
and supposing that #abs(2(x-1))<1# we get
#sum_{j=0}^{infty}(2(x-1))^j = 1/(1-2(x-1)) = 1/(3-2x)#
Putting all together
#(df)/(dx)(x) =2/(3-2x)# for #abs(x-1)<1/2#
integrating
#f(x)=int 2/(3-2x)dx = c_0-log_e(3-2x)#
but #f(1)=0->c_0=0#

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