I am not sure how I need to solve this. Can someone get me started?

How many terms of the following series would you need to add to find its sum to within 0.01?

$\displaystyle \sum^{\infty}_{n=2} \frac{4}{n(ln(n))^9}$

a) $\displaystyle n > e^{(25)^{\frac{1}{4}}}$

b) $\displaystyle n > e^{(50)^{\frac{1}{8}}}$

c) $\displaystyle n > e^{(100)^{\frac{1}{8}}}$

d) $\displaystyle n > e^{(25)^{\frac{1}{8}}}$

e) $\displaystyle n > e^{(100)^{\frac{1}{4}}}$