Show that if we want to approximate the sum of the series N_infinity n^-1.001 so that the error is less than 5 in the ninth decimal place, then we need to add more than 10^11.301 terms.
need help please.
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