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

1 Answer
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