If $f(s) = (1+s)^{(1+s)^{(1+s)/s}/s}$, show that $\lim_{s \to \infty} f(s)/s = 1$.

Question

If $f(s) = (1+s)^{(1+s)^{(1+s)/s}/s}$, show that $\lim_{s \to \infty} f(s)/s = 1$.

This function comes up in the parameterization of the solutions to $x^y = y^x$. See for example, here: Are there real solutions to $x^y = y^x = 3$ where $y \neq x$?

If we write $y = rx$, we get $x = r^{1/(r-1)}$ and $y=r^{r/(r-1)}$.

Then $x^y =(r^{1/(r-1)})^{r^{r/(r-1)}} =r^{r^{r/(r-1)}/(r-1)} $.

Finally, if we write $r = 1+s$, this is $f(s) = (1+s)^{(1+s)^{1+1/s}/s} $.

Wolfy says that, around $s=0$, $$f(s) =e^e + \dfrac{e^{1 + e} s^2}{24} - \dfrac{e^{1 + e} s^3}{24} + \dfrac{e^{1 + e} (219 + 5 e) s^4}{5760} + O(s^5) $$ and, around $s = \infty$, $$f(s) =s + (\log^2(1/s) - \log(1/s) + 1) + \dfrac{\log^4(1/s) - 3 \log^3(1/s) + 5 \log^2(1/s) - 6 \log(1/s) + 2}{2 s} + O((1/s)^2), $$ calling this a generalized Puiseux series.

My question is: How to show that that expansion at $s = \infty$ is correct. I would be satisfied with a proof that, as the title says, $\lim_{s \to \infty} \dfrac{f(s)}{s} = 1 $.

It might help to note that

$\dfrac{y}{x} = r$ and $\dfrac{r}{s} = \dfrac{1+s}{s} = 1+\dfrac1{s} $.

Answer 1

The expansion at infinity is actually easy. Following general methods to provide expansions at infinity:

$$f(1/s)=(1+1/s)^{(1+1/s)^{(1+1/s)/(1/s)}/(1/s)}=(1+1/s)^{(s+1)\cdot(1+1/s)^s}$$

As $s\to0^+$, we know that

$$(1+1/s)^s\to1$$

$$\implies(1+1/s)^{(s+1)\cdot(1+1/s)^s}\sim(1+1/s)^{(s+1)}$$

Which gives us

$$s\cdot f(1/s)\sim s\cdot(1+1/s)^{(s+1)}\to1$$

since

$$s\cdot(1+1/s)^{(s+1)}=\frac{s^{s+2}}{(s+1)^{s+1}}\to\frac11$$

coming from well known proofs of $\lim_{x\to0^+}x^x=1$.

Answer 2

$$\frac{f(s)}{s} = (1+s)^{(1+s)^{(1+s)/s}/s}/s$$ $$\log(\frac{f(s)}{s}) = \frac{(1+s)^{(1+s)/s}\log(1+s)}{s}-\log(s)$$ We are now claiming that $$\lim_{s \to \infty} \left(\frac{(1+s)^{(1+s)/s}\log(1+s)}{s}-\log(s)\right)=0$$

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We now see that the LHS can be changed into $$ \lim_{s \to \infty}\left(\frac{(1+s)^{(1+s)/s}\log(1+s)-s\log(s)}{s}\right)$$ Now is as good a time as ever to let $r=s+1$ $$ =\lim_{r \to \infty}\left(\frac{r^{\frac{r}{r-1}}\log(r)-(r-1)\log(r-1)}{r-1}\right)$$ $$ =\lim_{r \to \infty}\left(\frac{r\log(r)-(r-1)\log(r-1)}{r-1}\right)$$ By L'Hopital $$ =\lim_{r \to \infty}\left(\log(r)-\log(r-1)\right) = \color{red}{0}$$