Approximate the error of a alternating series

[rinspd]

determine the number of terms requires to approximate the sum of the series with an error less than 0.0001 and use a graphing utility to approximate the sum of the series with error less than 0.0001

Σ(lower: k=0, upper: infinity) (-3)^k/(1*3*5****(2k+1))

 

[HallsofIvy]

determine the number of terms requires to approximate the sum of the series with an error less than 0.0001 and use a graphing utility to approximate the sum of the series with error less than 0.0001

Σ(lower: k=0, upper: infinity) (-3)^k/(1*3*5****(2k+1))

Did you read the rules for this forum? You must show what you have tried yourself.

Notice that if k= 0, this is 1/3. If k= 1, it is -3/(1*3*5)=1/3 -1/5= 2/15= 0.1333. If k= 2 it is 2/15+ 9/(1*3*5*7)= 2/15+ 3/35= (14+ 9)/35= 23/35= 0.6571 . Do you see that each sum lies between the previous two sums? Do you see why that is?