The series $\backslash$sum$\_($n$=1)^(\backslash$infty $)((-1)^($n$+1))/(2^($n$))$s $6^($th $)$ partial sum $($rounded to $4$ decimal places$)$:

$\backslash$sum$\_($n$=1)^(\backslash$infty $)((-1)^($n$+1))/(2^($n$))$~~s$\_(6)$~~

$($rounded to $4$ decimal places$)$

According to the Alternating Series Remainder Theorem, the error involved in our approximation is bounded

by:

$|$s$-$s syntax incomplete. $($exact value$)$ s$\_(6)$ and the maximum errors is guaranteed to land. Give your answer in interval notation and

round the endpoints to $4$ decimal places:

Interval containing s :