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:

  1. Using an intuitive or geometric explanation, why doesn't $ 1\over x $ converge to some value?
  2. Why is the volume described above finite while the cross-section is infinite?

Edit: Changed $[]$ to $()$

  • 4
    $\begingroup$ Unfortunately, intuitition is not very helpful here. It is difficult to visualize whether a function tends to $0$ fast enough. Take for example $\frac{1}{x^{1.00001}}$. I doubt you can distinguish it visually from $\frac{1}{x}$ $\endgroup$
    – Peter
    Commented Feb 8, 2018 at 20:50
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    $\begingroup$ I noted that as well. Which only raises my curiosity even further. What's the "difference" (for a lack of a better term) between $ 1/x $ and $ 1/x^{1.000 ... 0001} $? Is the difference in area infinite? I struggle to understand that. $\endgroup$
    – npengra317
    Commented Feb 8, 2018 at 20:53
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    $\begingroup$ Some folks use that notation for floor or ceiling. $\endgroup$
    – copper.hat
    Commented Feb 8, 2018 at 20:55
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    $\begingroup$ On the other hand, @npengra317, you have no problem visualising the 2D object having infinite width but finite area like the one below the curve $\frac{1}{x^2}$? I guess this should violate intuition in exactly the same way - yet we have all got used to it via years of maths' training. $\endgroup$
    – user491874
    Commented Feb 8, 2018 at 20:56
  • 2
    $\begingroup$ This is known as Gabriel's horn; or also, i believe, as Torricelli's trumpet... $\endgroup$
    – user403337
    Commented Feb 9, 2018 at 0:03

2 Answers 2


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.


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.


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