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.)