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

May 25, 2016

${\sum}_{i = 1}^{\infty} \frac{{2}^{i} {\left(x - 1\right)}^{i}}{i} = {\log}_{e} \left(\frac{1}{3 - 2 x}\right)$ for $\left\mid x - 1 \right\mid < \frac{1}{2}$

#### Explanation:

Let $f \left(x\right) = {\sum}_{i = 1}^{\infty} \frac{{2}^{i} {\left(x - 1\right)}^{i}}{i}$
calculating
$\frac{\mathrm{df}}{\mathrm{dx}} \left(x\right) = {\sum}_{i = 1}^{\infty} \left({2}^{i} {\left(x - 1\right)}^{i - 1}\right)$ simplifying
$\frac{\mathrm{df}}{\mathrm{dx}} \left(x\right) = 2 {\sum}_{j = 0}^{\infty} {\left(2 \left(x - 1\right)\right)}^{j}$
following the polynomial identity
$\frac{1 - {z}^{n + 1}}{1 - z} = 1 + z + {z}^{2} + {z}^{3} + \ldots + {z}^{n}$
and supposing that $\left\mid 2 \left(x - 1\right) \right\mid < 1$ we get
${\sum}_{j = 0}^{\infty} {\left(2 \left(x - 1\right)\right)}^{j} = \frac{1}{1 - 2 \left(x - 1\right)} = \frac{1}{3 - 2 x}$
Putting all together
$\frac{\mathrm{df}}{\mathrm{dx}} \left(x\right) = \frac{2}{3 - 2 x}$ for $\left\mid x - 1 \right\mid < \frac{1}{2}$
integrating
$f \left(x\right) = \int \frac{2}{3 - 2 x} \mathrm{dx} = {c}_{0} - {\log}_{e} \left(3 - 2 x\right)$
but $f \left(1\right) = 0 \to {c}_{0} = 0$