Distance between the nullpoints of the series of derivatives of ln(x)/x I plotted a function $f(x) = \frac{ln(x)}{x}$, and continued with $f'(x)$, $f''(x)$, $f'''(x)$. I noticed how the intersections between the functions and the x-axis seemed to be roughly equally distant apart:

It turned out they were not. However, the distance seems to converge. After a few iterations:
1.71828
1.76341
1.77301
1.77649
1.77813
1.77992
1.78016
1.78033  
The derivatives follow a simple pattern:

The exponent of the divisor increases by one for each increment, the number before the $ln(x)$ is the factorial function, and the constant term apparently belongs to unsigned Stirling numbers of the first kind
I am just curious, what does this converge into? The only way I can solve it is to brute-force it, and then guess a closed form, or use wolfram alpha. Nothing useful pops up at 1.78033 though.
 A: It can be shown that
$$ f^{(n)}(x) = (-1)^n {n! \log x - a(n) \over x^n} $$
where $a(n) = s(n+1, 2)$ is as given in https://oeis.org/A000254 - note that there's a recurrence given here, $a(n+1) = (n+1) a(n) + n!$, which could be used to prove this (for example by induction).
Therefore $ f^{(n)}(x) = 0 $ when $n! \log x = a(n)$, i. e. when $x = \exp(a(n)/n!)$.  One of the notes to that sequence gives $a(n) = n! H(n)$, where $H(n) = 1 + 1/2 + ... + 1/n$ is the $n$th harmonic number.  So you have that $f^{(n)}(x) = 0$ when $x = \exp H_n$.  Call this $x_n$.
Then you are interested in $\lim_{n \to \infty} x_n - x_{n-1}$, if it exists.  We have
$$ x_n - x_{n-1} = \exp(H_n) - \exp(H_{n-1}) = \exp(H_n) (1 - e^{-1/n}) $$
and we have the asymptotic series for the factors:
$$ \exp(H_n) = \exp (\log n + \gamma + O(1/n)) = n e^\gamma (1 + O(1/n)) $$
where $\gamma$ is the Euler-Mascheroni constant (https://en.wikipedia.org/wiki/Euler–Mascheroni_constant),  and
$$ 1 - e^{-1/n} = 1 - \left( 1 - 1/n + O(1/n^2) \right) = 1/n + O(1/n^2) = 1/n (1 + O(1/n)) $$
Multiplying these together we get
$$ \exp(H_n) (1 - e^{-1/n}) = \left( n e^\gamma (1 + O(1/n)) \right) \cdot  {1 \over n} (1+O(1/n)) $$
or, after mmultiplication, $e^\gamma (1 + O(1/n))$.  Thus as $n \to \infty$ we have $\lim_{n \to \infty} x_n - x_{n-1} = e^\gamma$.  The constant $e^\gamma$ is somewhat important and in decimal this is about $1.78107$, which explains why you didn't find anything around $1.78033$ - you just didn't have a good enough approximation.
