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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.

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    $\begingroup$ @AndrewLi Wouldn't it be $\log 2+\log\log 3 +\log\log\log 4....?$ $\endgroup$
    – saulspatz
    May 9, 2018 at 2:08
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    $\begingroup$ @AndrewLi $\log(2\log 3)= \log 2 + \log\log 3$ Note that $\log 2 + \log 3 = \log 6$ $\endgroup$
    – saulspatz
    May 9, 2018 at 2:12
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    $\begingroup$ 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, 2018 at 2:17
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    $\begingroup$ @ChristopherMarley log(2)+log(log(3)+log(log(4)+...)))) $\endgroup$ May 9, 2018 at 2:18
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    $\begingroup$ 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, 2018 at 2:23

4 Answers 4

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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, $$\ln (2 \ln (3 \ln (4 \ln (5\cdots )))) \leq 2.$$

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    $\begingroup$ You want $e^{n}/n>n+1$ for the weakening steps. $\endgroup$ May 9, 2018 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". $\endgroup$ May 9, 2018 at 2:56
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    $\begingroup$ @EnricoBorba $x<y$ is implied by $z<y$ and $x < z$. $\endgroup$ May 9, 2018 at 2:59
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    $\begingroup$ @EnricoBorba I've made some edits to the question that should address any similar misunderstandings. $\endgroup$ May 9, 2018 at 3:14
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    $\begingroup$ @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, 2018 at 3:58
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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.

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  • $\begingroup$ Elegant approach and solution. $\endgroup$ May 9, 2018 at 4:03
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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.

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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.

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