What is the interval of convergence of sum_1^oo (2^k)/k (x-1)^k 12kk(x1)k?

1 Answer
May 25, 2016

sum_{i=1}^{infty}(2^i(x-1)^i)/i = log_e (1/(3-2x))i=12i(x1)ii=loge(132x) for abs(x-1)<1/2|x1|<12

Explanation:

Let f(x)=sum_{i=1}^{infty}(2^i(x-1)^i)/if(x)=i=12i(x1)ii
calculating
(df)/(dx)(x) = sum_{i=1}^{infty}(2^i(x-1)^{i-1})dfdx(x)=i=1(2i(x1)i1) simplifying
(df)/(dx)(x) = 2sum_{j=0}^{infty}(2(x-1))^jdfdx(x)=2j=0(2(x1))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