12
$\begingroup$

I was playing around with nested radicals and I decided to see if nested equations of logarithms would converge.

It seems to converge to a value around $1.368$, and at a depth of 20 it has a value of $1.3679012...$, however I am not sure how to prove whether it actually does converge.

$\endgroup$
  • 4
    $\begingroup$ @AndrewLi Wouldn't it be $\log 2+\log\log 3 +\log\log\log 4....?$ $\endgroup$ – saulspatz May 9 '18 at 2:08
  • 2
    $\begingroup$ @AndrewLi $\log(2\log 3)= \log 2 + \log\log 3$ Note that $\log 2 + \log 3 = \log 6$ $\endgroup$ – saulspatz May 9 '18 at 2:12
  • 12
    $\begingroup$ Fellas, $$\begin{array}{ll} \ln(2\ln(3\ln(4))) & =\ln(2)+\ln\ln(3\ln4) \\ & =\ln(2)+\ln(\ln3+\ln\ln4) \\ & \color{Red}{\ne}\ln(2)+\ln\ln3+\ln\ln\ln4\end{array}$$ $\endgroup$ – anon May 9 '18 at 2:17
  • 2
    $\begingroup$ @ChristopherMarley log(2)+log(log(3)+log(log(4)+...)))) $\endgroup$ – Hagen von Eitzen May 9 '18 at 2:18
  • 2
    $\begingroup$ anon Certainly true. However, for $A,B>2$ we have $\ln(A+B) < \ln(A) + \ln(B)$. So if @saulspatz's expression converges, this one will too. $\endgroup$ – mweiss May 9 '18 at 2:23
12
$\begingroup$

For all integers $n \geq 2$, note that $$n(n+1) < e^n. \tag{*}$$ We may show this by expanding $e^n > 1 + n + \frac{n^2}{2} + \frac{n^3}{6}$, which implies: $$e^n- n(n+1) > 1 - \frac{n^2}{2} + \frac{n^3}{6} = \left(\frac{n}{6} - \frac{1}{2}\right)n^2 + 1.$$ The RHS is manifestly positive for $n \geq 3$ and can be checked to be positive for $n = 2$.

Now, starting with $\ln n < n$ (which should need no proof), multiply on both sides by $n-1$ and apply $(\text{*})$ to show that $$(n-1) \ln n < (n-1) n < e^{n-1}.$$ Take logarithms to get $$\ln ( (n-1) \ln n) < n-1,$$ then multiply on both sides by $n-2$ to show that $$(n-2) \ln ( (n-1) \ln n) < (n-2) (n-1) < e^{n-2},$$ or $$\ln ((n-2) \ln ( (n-1) \ln n)) < n-2.$$ Multiply by $n-3$, apply $(\text{*})$, and take logarithms again to get $$\ln ((n-3) \ln ((n-2) \ln ( (n-1) \ln n))) < n-3.$$ By proceeding similarly, we can show that for $2 \leq k < n$ arbitrary, $$\ln (k \ln ((k+1) \cdots \ln n)) < k.$$ The LHS of this inequality is monotone increasing and bounded above as $n \to \infty$, so it must have a limit $$\ln (k \ln ((k+1) \ln ((k+2) \cdots))) \leq k.$$ In particular, $$\ln (2 \ln (3 \ln (4 \ln (5\cdots )))) \leq 2.$$

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ You want $e^{n}/n>n+1$ for the weakening steps. $\endgroup$ – RideTheWavelet May 9 '18 at 2:44
  • $\begingroup$ Maybe I am just utterly lost but, If you have, x < y and z < y, why would that imply x < z? You seem to do this for every step of the "weakening". $\endgroup$ – Enrico Borba May 9 '18 at 2:56
  • 2
    $\begingroup$ @EnricoBorba $x<y$ is implied by $z<y$ and $x < z$. $\endgroup$ – Connor Harris May 9 '18 at 2:59
  • 1
    $\begingroup$ @EnricoBorba I've made some edits to the question that should address any similar misunderstandings. $\endgroup$ – Connor Harris May 9 '18 at 3:14
  • 1
    $\begingroup$ @CMonsour but: $\ln$ is monotonic increasing after 0 and $x\ln(x+1)>x$ for $x\ge 2$, So it is monotonic increasing, add the bounded from above and you get? $\endgroup$ – ℋolo May 9 '18 at 3:58
8
$\begingroup$

For $n\le m$, let $$ f(n,m)=n\ln((n+1)\ln(\ldots (m)\ldots))$$ i.e., $$f(n,m)=\begin{cases}n&n=m\\n\ln(f(n+1,m))&n<m\end{cases} $$ We want $\lim_{m\to\infty}f(1,m)$.

Clearly, $f(n,\cdot)$ is increasing (in particular, $f(n,m)\ge n$) so that convergence equals boundedness. Compare $f(n+1,m+1)$ against $f(n,m)$. If $m=n>10$, $f(n+1,m+1)=f(n,m)+1<2f(n,m)$. By induction on $m-n$, for $m>n\ge 10$ as well $$ \begin{align}f(n+1,m+1)&=(n+1)\ln( f(n+2,m+1) )\\&<(n+1)\ln(2f(n+1,m))\\&=(n+1)(\ln 2+\ln(f(n+1,m))\\&<(1+\tfrac1{10})n\cdot (1+\tfrac{\ln2}{\ln11})\ln(f(n+1,m))\\&<2f(n,m)\end{align}$$ So $$f(n,m)<f(n+1,m+1)<2f(n,m)\qquad m\ge n\ge 10 $$ This makes $$\tag1f(n,m)<f(n,m+1)=n\ln f(n+1,m+1)<n\ln f(n,m)+n\ln 2 $$ for $n\ge 10$. The right hand side is slower than linear in $f(n,m)$, hence $f(n,m)$ is bounded from above, $\lim_{m\to\infty}f(n,m)$ exists and ultimately so does $\lim_{m\to\infty}f(1,m)$

Remark: Numerically, $(1)$ gives us $f(10,m)<44.998$. This trickles down to an upper bound $$f(1,m)< 1.36794$$ But similarly, we find $f(20,m)<107$ and with that can improve the bound to $$f(1,m)<1.3679012618$$ (for comparison, $f(1,20)>1.3679012615$). Starting with a bound for $f(50,m)$, we can compute $$\lim f(1,m)=1.367901261797085169668909175760\ldots$$ to 30 decimals.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Elegant approach and solution. $\endgroup$ – Claude Leibovici May 9 '18 at 4:03
0
$\begingroup$

Define for $n\geq 1$ the numbers $x_{n+1}=\exp(x_{n}/n)$. Then $x_{1}=\log(x_{2})=\log(2\log(x_{3})),$ and so on.

We may observe that if $x_{1}\leq 1.367,$ then the sequence $x_{n}$ initially increases, then quickly decreases and stabilizes. If the $x_{n}$ are actually bounded, then $\lim_{n\rightarrow\infty}\exp(x_{n}/n)=1$, so we should have $x_{n}\rightarrow 1$. Also, we can see that if for some $n\geq 3,$ $x_{n}<n,$ then $x_{n+k}<e$ for all $k\geq 1$, so this is a sufficient condition for the sequence to stabilize (and therefore converge to 1).

If $x_{1}>1.368,$ then this sequence very quickly blows up. If for some $n\geq 23,$ $x_{n}>n^{3/2},$ then $x_{n+1}\geq\exp(\sqrt{n})>(n+1)^{3/2},$ so the sequence grows exponentially. This can be improved to a $n^{1+\epsilon}$ bound if $n$ is taken large enough, since $\exp(n^{\epsilon})>(n+1)^{1+\epsilon}$ eventually.

So I conjecture that the value of this nested log has a collection of lower bounds given by $x_{1}$ for which the sequence $\{x_{n}\}$ eventually stabilizes, and a collection of upper bounds given by $x_{1}$ for which the sequence blows up. My guess is that the closer $x_{1}$ is to the true value, the further along in the sequence one will need to go to see the stabilization or blowup behavior.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Here's a Mathematica code, which might be helpful.

It is slightly strange as it starts on the top and ends on the bottom. Maybe not very elegant but it works.

Part 1: Check with small numbers: symbolic expressions and numeric values

 With[{digits = 10, nmax = 5, finalnums = 3},
 Table[ClearAll[f];
  With[{k = n}, f[k] = (k - 1) Log[k]];
  f[n_] := (n - 1) Log[f[n + 1]];
  {f[2], N[f[2], digits]}, {n, nmax - finalnums + 1, nmax}]]

Out[1093]= {
{Log[2 Log[3]], 0.7871950082}, 
{Log[2 Log[3 Log[4]]], 1.047491996}, 
{Log[2 Log[3 Log[4 Log[5]]]], 1.235680628}
}

Part 2: Calculation with high numerical precision

With[{digits = 40, nmax = 65, finalnums = 7},
 Table[ClearAll[f];
  With[{k = n}, f[k] = (k - 1) Log[k]];
  f[n_] := (n - 1) Log[f[n + 1]];
  N[f[2], digits], {n, nmax - finalnums + 1, nmax}]]

Out[1058]=    
{
1.367901261797085169668909175760488538295, \
1.367901261797085169668909175760488538369, \
1.367901261797085169668909175760488538382, \
1.367901261797085169668909175760488538384, \
1.367901261797085169668909175760488538385, \
1.367901261797085169668909175760488538385, \
1.367901261797085169668909175760488538385, \
1.367901261797085169668909175760488538385
}

You can see the convergence in the last lines of the output.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.