ln(1.01) and 2/201 can be thought of as evaluating ln(1 + 1/x) and 1/(x + 1/2) at x=100. Then if ln(1 + 1/x) is greater than 1/(x + 1/2) then [ln(1 + 1/x)]/[1/(x + 1/2)] > 1.
lim x->inf [ln(1 + 1/x)]/[1/(x + 1/2)] can be evaluated by first applying L'Hopital's rule (and simplify a bit) to get
lim x->inf(x^2+x+1/4)/(x^2+x)
This is pretty easy to get the limit equals 1.
d/dx [ln(1 + 1/x)]/[1/(x + 1/2)] = (x + 1/2)*ln(1 + 1/x) comes out pretty straightforward as
ln(1 + 1/x) - (x + 1/2)/(x^2+x)
This can be manipulated into ln(1 + 1/x) - 1 - 4/(x^2+x)
Now, for any x>=1, 1<1+1/x<=2<e, which bounds ln(1 + 1/x) between 0 and 1. The important consequence of this is that d/dx will then always be negative for x>=1.
So now on [1,inf) we have [ln(1 + 1/x)]/[1/(x + 1/2)] is continuously decreasing and has a horizontal asymptote of 1. This means that the function must be greater than 1 for all x>=1.
So then f(100) = ln(1.01)/(2/201) > 1, which means that ln(1.01) > 2/201.
Solution
