Question

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

Answer 1

`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`