# 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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embizze | Certified Educator

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.)

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