
Yefim S. answered 12/08/23
Math Tutor with Experience
Last quetion>
1/[n2n) < 0.0001;
It started from n = 10. So, n ≥ 10.
Chernobog S.
asked 12/07/23Test the series for convergence or divergence, I already solved all of it except the last part
∑(n=1, infinity) ((-1)^n)/((n)(2^n))
bn=1/(n2^n)
lim as n approaches infinity bn= 0; The series is convergent
If the series is convergent, use the Alternating Series Estimation Theorem to determine how many terms we need to add in order to find the sum with an error less than 0.0001? (If the quantity diverges, enter DIVERGES.) How many terms?
Yefim S. answered 12/08/23
Math Tutor with Experience
Last quetion>
1/[n2n) < 0.0001;
It started from n = 10. So, n ≥ 10.
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