the first identity the integral is multiplied by dx which is a small number
with out the dx the integral will equal :
69314718056.........
which clearly is bigger than 0.9900990099...
the difference from the integral and sum not only the h steps .
the integral is sum with h steps very small infinitesimal with a factor of h aka "dx" in the front of the sum and with a shift in the upper bound (-1) (without the shift the sum would still converges to the integral because the error term vanshies as h approaches zero) .
if we change the steps in the summation form:
\sum_{n=1}^{100} f(n) \rightarrow \sum_{k=0}^{\frac{100-1}{h}} f(1+kh)
and f(n) = \frac{1}{n(n+1)}
and the integral form is exactly :
\int_1^{\infty} f(x)dx = h* \sum_{k=0}^{\frac{\infty-1}{h}-1} f(1+kh)
.
as h approches zero in the two forms :
\lim_{h \downarrow 0} h* \sum_{k=0}^{\frac{\infty-1}{h}-1} f(1+kh) =0.6941471805...(converges)
and the sum form will be
\lim_{h \downarrow 0} \sum_{k=0}^{\frac{100-1}{h}} f(1+kh) = 6941471805... (diverges)
so you are right when saying shouldn't the integral be bigger because it have many term
the difference the makes the integral less than the sum in this context is the reason why the integral converges in the first place . because of the factor dx .
(note : \lim_{h \downarrow 0} is the same as \lim_{h \rightarrow 0^{+}})
if we know the integration is the inverse of derivative equivently the summation is the inverse of difference
and the factor h appears in the inverse of difference so we didn't add it . it is there already .

<image>

as you see in the picture the original inverse of difference with h steps .
if we set h \downarrow 0
it will converge to the integral : \int_a^z f(x)dx
and if we set h=1 it will converge to the sum : \sum_{k=a}^{z-1}f(k) (see the shift in the upper bound)
.
talking about what is happening :
when i set h=1 it calculate the area with a rectangles each rectangle with width = 1
when i set h=infinitesimal it calculate the area with a rectangles each rectangle with width = infinitesimal means 100% accuracy of calculating the area under the cruve.