Expected value of a discrete random variable with $R(X)$ countably infinite

Question

Consider a discrete random variable with pdf $f(x_k)=\displaystyle\frac{1}{2^k}$ with $R(X)$ $\displaystyle\{x_k =(-1)^k\frac{2^k}{k},k=1,2,3...\}$ ,then the sum defining the expectation is $$\sum_{k=1}^{\infty }{x_kf(x_k)}=\sum_{k=1}^{\infty }{\frac{(-1)^k}{k}}=-ln(1+1)=-ln(2)$$

Can someone explain how this sum converges to $-ln(1+1)$ ?

Answer 1

Taylor series expansion of $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...$. Putting $x=1$, we get $\ln(2)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=-\sum_{k=1}^{\infty}\frac{(-1)^k}{k}$. Hence $\sum_{k=1}^{\infty}\frac{(-1)^k}{k}=-\ln(2)$.

Answer 2

It looks like $$\sum_{k = 1}^\infty\frac{(-1)^k}{k}x^k = (-1)\sum_{k = 1}^\infty\frac{(-1)^{k+1}}{k}x^k = -\log (1+x).$$ When $x = 1$, this is equal to $-\log (1+1) = -\log (2).$