It is my opinion that this problem would require something other than adding the given terms out sufficiently far for 14-digit accuracy. The following inequality is obviously true:
int(1/x^s,n,n+1) > 1/(n+1)^s > int(1/x^s,n+1,n+2)
which means that the infinite sum
1/(n+1)^s + 1/(n+2)^s + 1/(n+3)^s +...
lies between int(1/x^s,n,inf) and int(1/x^s,n+1,inf). These two quantities can be solved as (s-1)/n^(s-1) and (s-1)/(n+1)^(s-1). With the s = 1.1 value, to ensure that the sum which is left out,
1/(n+1)^s + 1/(n+2)^s + 1/(n+3)^s +...
is not more than 10^(-14) it we would have to have
(s-1)/(n+1)^(s-1) < 10^(-14),
or substituting s = 1.1
.1/(n+1)^.1 < 10^(-14)
10^13 < (n+1)^.1
10^130 < n+1
The number 10^130 is horribly large and the consequent cumulative round off error would be enormous, though the universe would suffer a heat death before the computation was completed. However, if we get smart, we can use the above integral to complete the summation. Choose n sufficiently large that the difference between the two integrals is less than 10^(-14):
(s-1)*(1/n^(s-1)-1/(n+1)^(s-1)) < 10^(-14)
1/n^.1-1/(n+1)^.1 < 10^(-13)
The answer to that is n = 81114515936, which above 81 billion. The plan would be then, for s = 1.1, to add the first n = 81114515936 terms, but in reverse order to minimize accumulated rounding error, and then add on the value .1/81114515936^.1 for the integral.
This is a far cry from stopping after a couple million terms. However, in any case, the cumulative rounding error from 81 billion additions is likely to greatly exceed 10^(-14) which indicates in my mind, a hastily conceived homework problem.
