## anonymous 5 years ago Evaluate the divergent series, or state that it diverges 7*∑(k=1 to ∞) ((-1)^(k-1)/2^(k-1))

1. anonymous

$7\times \sum_{k=1}^{\infty}((-1)^\left( k-1 \right))/(2^\left( k-1\right))$

2. anonymous

Fortunately, this series is already in the form of a geometric series, i.e. $\sum_{n=1}^{\infty}ar^{n-1}=a/(1-r)$ and these series are convergent when $\left| r \right|<1$ So, saying that: $7*\sum_{k=1}^{\infty}(-1/2)^{k-1},r=-1/2,a=1$$7*\sum_{k=1}^{\infty}(-1/2)^{k-1}=7/(1+(1/2))=7/(3/2)=14/3$ The above is the value of the series, however you know it is convergent because r=-1/2 and $\left| r \right| = \left| -1/2 \right|=1/2 < 1$