What is this sequence that converges to ln(x) called?

  • Context: Undergrad 
  • Thread starter Thread starter Daniel Gallimore
  • Start date Start date
  • Tags Tags
    Sequence
Click For Summary
SUMMARY

The convergent sequence for the natural logarithm is defined as \(\lim_{a\rightarrow\infty} a x^{1/a}-a=\ln(x)\). This sequence, along with the Hurwitz sequences \(a_{n}=2^{n}(x_{n}-1)\) and \(b_{n}=2^{n}(1-1/x_{n})\), converges to \(\ln(x)\). The equivalence of \(\ln x\) can also be expressed through limits involving \(n\) and \(h\), as shown in the discussion. These sequences highlight various approaches to defining the natural logarithm through limits and integrals.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with sequences and series
  • Knowledge of L'Hôpital's rule
  • Basic concepts of integrals and derivatives
NEXT STEPS
  • Research the properties of Hurwitz sequences
  • Explore the derivation of logarithmic functions using limits
  • Study the applications of L'Hôpital's rule in calculus
  • Learn about the relationship between sequences and their convergence
USEFUL FOR

Mathematicians, calculus students, and anyone interested in advanced mathematical concepts related to logarithmic functions and their properties.

Daniel Gallimore
Messages
48
Reaction score
17
I found the following convergent sequence for the natural logarithm online: \lim_{a\rightarrow\infty}a x^{1/a}-a=\ln(x) Does anybody know where this sequence first appeared, or if it has a name?
 
Physics news on Phys.org
There is an attempt to list known integer sequences: http://oeis.org/ but I doubt that this one has a certain name. Alfred Hurwitz has found these two sequences with ##\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = \ln x##
$$\begin{aligned}
a_{n}&=2^{n}(x_{n}-1)\\b_{n}&=2^{n}(1-1/x_{n})\end{aligned}$$
which thus could be called Hurwitz sequences. Wikipedia says (which means I haven't checked) ##\ln x = \lim_{n \to \infty} n \left( 1 - \frac{1}{\sqrt[n]{x}} \right) ## is equivalent to ## \ln x= \lim_{h \to 0} \frac{x^{h}-1}{h} = \lim_{h \to 0} \int_{1}^{x}\frac{1}{t^{1-h}}\, dt## by L’Hôpital's rule. So the sequence you mentioned is in a way the natural definition, given that ##\ln |x| = \int \frac{1}{x}\,dx##
 
The limit statement is equivalent to
$$\lim_{h\to 0} \frac{c^h-1}h=\ln c$$
which gives the derivative of ##c^x## at ##x=0##.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K