sum_{n=1 to infty} 1/n^1.001 = sum_{n=1 to N} 1/n^1.001 + sum_{n=N+1 to infty) 1/n^1.001,

but:

int_{x=N1 to infty} 1/x^1.001 dx < sum_{n=N+1 to infty) 1/n^1.001 < int_{x=N to infty} 1/x^1.001 dx

so if we choose N so that int_{x=N to infty} 1/x^1.001 dx <= 0.000000005

then we are done. The integral is 1000/N^0.001, so we seek N such that:

5 10^-9 = 1000/N^0.001

which has solution:

N ~= 1.07 10^11301

which is more than 10^11301 terms.

RonL