We are asked to determine the value for n such that the indicated sum is correct to 1 decimal place.
S_n=sum_(n=1)^(oo) n/(n^2+n^3)
The series converges (using the comparison test) and the sum is 1. The formula for the partial sum is
sum_(n=1)^m n/(n^2+n^3)=m/(m+1)
To be accurate to one decimal place we need the partial sum to be greater than or equal to .95:
m/(m+1)=.95 ==> .05m=.95 ==> m=19
Thus we need 19 terms for the partial sum to be accurate to one decimal place.
The required value is n=19.
(For n=19 we have the sum of the series equal to .95 which is 1.0 to one decimal place. for n=18 the sum is approximately 0.9473684211 which is 0.9 to one decimal place.)
http://mathworld.wolfram.com/ComparisonTest.html
Thursday, April 16, 2015
For what value of n will sn be correct to 1 decimal place? Use s10=10/11 to estimate and compare the sum sum_(n=1)^oo n/(n^2+n^3)
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