Question

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

Answer 1

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

Answer 2

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 \[\left| r \right| = \left| -1/2 \right|=1/2 < 1\]