Geometric (or Intuitive) proof of the improper integration of $\frac1x $ From a mathematical standpoint, I understand and I can solve the following:
$$ \lim_{M\rightarrow\infty} \int_1^M \left({1 \over x}\right) \rightarrow \infty $$
Additionally,
$$ \lim_{M\rightarrow\infty} \int_1^M \left({1 \over x^2}\right) \rightarrow 1 $$
This all makes mathematical sense to me. It's the geometric parts that confuse me. 
The family of of $ 1/x^p $ graphs look very similar to me, so it makes me wonder why $ 1/x $ doesn't converge to some value as well.
Especially considering the fact when you rotate $ 1/x $ and calculate the volume of that shape; it converges to some value. Again, mathematically, this makes sense because:
$$ \lim_{M\rightarrow\infty}\int_1^M \left({1 \over x}\right) dx > \lim_{M\rightarrow\infty} \int_1^M \pi\left({1 \over x^2}\right) dx $$
But the geometric implications of this are that a cross-section of such an object has an infinite area but the volume is some finite value.
My questions:


*

*Using an intuitive or geometric explanation, why doesn't $ 1\over x $ converge to some value?

*Why is the volume described above finite while the cross-section is infinite?


Edit: Changed $[]$ to $()$
 A: Here's something that may be helpful geometrically, borrowed from a classic argument about the harmonic series.
One argument for the divergence of the harmonic series goes as follows:
$$ \begin{align}
&\frac12 + \frac13 + \frac14 + \frac15 + \frac16 + \frac17 + \frac18 + \dots \\
> &\frac12 + \frac14 + \frac14 + \frac18+ \frac18+ \frac18+ \frac18 + \dots \\
= &\frac12 + \left(\frac14 + \frac14\right) + \left(\frac18+ \frac18+ \frac18+ \frac18\right) + \dots \\
= &\frac12 + \frac12 + \frac12 + \dots
\end{align} $$
Let's translate this into our integral. Imagine the area under the curve $\frac1x$. Fit a rectangle of area $\frac12$ between the points $(1,0),(2,0),(1,\frac12),(2,\frac12)$ -- this inside the area of the curve. Fit the next rectangle of area $\frac12$ between the points $(2,0),(4,0),(2,\frac14),(4,\frac14)$. In general the $i$th rectangle will be placed between the points $(2^{i-1},0),(2^i,0),(2^{i-1},\frac1{2^i}),(2^i,\frac1{2^i})$. No two rectangles overlap, and each rectangle has area $\frac12$. Since you can fit infinite rectangles of equal area under the integral, it must diverge.
A: Suppose
$\lim_{M\rightarrow\infty} \int_1^M \dfrac{dx}{x}
$
exists.
Letting $L$
be this limit,
$\lim_{M\rightarrow\infty} \int_1^M \dfrac{dx}{x}
=L$
so that,
for any $c>0$
there is a $M(c)$
such that
$0 \lt L-\int_1^M \dfrac{dx}{x}
\lt c$
for $M > M(c)$.
Choosing such an $M$,
we also have
$0 
\lt L-\int_1^{2M} \dfrac{dx}{x}
\lt c$
so that
$0
\lt L-\int_1^{2M} \dfrac{dx}{x}
=L-\int_1^{M} \dfrac{dx}{x}-\int_M^{2M} \dfrac{dx}{x}
$
or
$\int_M^{2M} \dfrac{dx}{x}
\lt L-\int_1^{M} \dfrac{dx}{x}
\lt c$.
But
$\int_M^{2M} \dfrac{dx}{x}
\gt \dfrac{M}{2M}
=\dfrac12$.
This is a contradiction
for $c < \dfrac12$.
This is, of course,
a restating of
the standard elementary proof
that the harmonic sum diverges.
