Answer
(a) and (c) are equivalent; (b) is not equivalent to the other two.
Work Step by Step
(a)
4$\Sigma_{k=2}$ $\frac{(-1)^{k-1}}{k-1}$= $\frac{(-1)^{2-1}}{2-1}$+ $\frac{(-1)^{3-1}}{3-1}$+ $\frac{(-1)^{4-1}}{4-1}$ =-1+(1/2)-(1/3)
(b)
2$\Sigma_{k=0}$ $\frac{(-1)^{k}}{k-1}$= $\frac{(-1)^{0}}{0+1}$+ $\frac{(-1)^{1}}{1+1}$+ $\frac{(-1)^{2}}{2+1}$ =1-(1/2)+(1/3)
(c)
1$\Sigma_{k=-4}$ $\frac{(-1)^{k}}{k+2}$= $\frac{(-1)^{-1}}{-1+2}$+ $\frac{(-1)^{0}}{0+2}$+ $\frac{(-1)^{1}}{1+2}$ =-1+(1/2)-(1/3)
Hence proved
(a) and (c) are equivalent; (b) is not equivalent to the other two.
