Jtdylktuy

Suppose $x_n$ is the only positive solution to the equation $x^{-n}=sumlimits_{k=1}^infty (x+k)^{-n}$,how to show the existence of the limit $lim_{nto infty}frac{x_n}{n}$?

It is easy to see that ${x_n}$ is increasing.In fact, the given euation equals
$$1=sum_{k=1}^infty(1+frac{k}{x})^{-n} tag{*}$$
If $x_nge x_{n+1}$,then notice that for any fixed$ k$,$(1+frac{k}{x})^{-n}$ is increasing,thus we can get
$$frac{1}{(1+frac{k}{x_n})^n}ge frac{1}{(1+frac{k}{x_{n+1}})^n}>frac{1}{(1+frac{k}{x_{n+1}})^{n+1}}$$
By summing up all k's from 1 to $infty$,we can see
$$sum_{k=1}^inftyfrac{1}{(1+frac{k}{x_n})^n}>sum_{k=1}^inftyfrac{1}{(1+frac{k}{x_{n+1}})^{n+1}}$$
then from $(*)$ we see that the two series in the above equality are all equals to $1$,witch is a contradiction!

But it seems hard for us to show the existence of $lim_{nto infty}frac{x_n}{n}$.What I can see by the area's principle is

$$Big|sum_{k=1}^inftyfrac{1}{(1+frac{k}{x_n})^n}-int_1^infty frac{1}{(1+frac{x}{x_n})}dxBig|<frac{1}{(1+frac1{x_n})^n}$$
or
$$Big|1-frac{x_n}{n-1}(1+frac{1}{x_n})^{1-n}Big|<frac{1}{(1+frac1{x_n})^n}$$

For any $n ge 2$, consider the function $displaystyle;Phi_n(x) = sum_{k=1}^infty left(frac{x}{x+k}right)^n$.

It is easy to see $Phi_n(x)$ is an increasing function over $(0,infty]$.
For small $x$, it is bounded from above by $x^n zeta(n)$ and hence decreases to $0$ as $x to 0$.
For large $x$, we can approximate the sum by an integral and $Phi_n(x)$ diverges like $displaystyle;frac{x}{n-1}$ as $x to infty$. By definition, $x_n$ is the unique root for $Phi_n(x_n) = 1$. Let $displaystyle;y_n = frac{x_n}{n}$.

For any $alpha > 0$, apply AM $ge$ GM to $n$ copies of $1 + frac{alpha}{n}$ and one copy of $1$, we obtain

$$left(1 + frac{alpha}{n}right)^{n/n+1} > frac1{n+1} left[nleft(1 + frac{alpha}{n}right) + 1 right] = 1 + frac{alpha}{n+1}$$
The inequality is strict because the $n+1$ numbers are not identical. Taking reciprocal on both sides, we get
$$left( frac{n}{n + alpha} right)^n ge left(frac{n+1}{n+1 + alpha}right)^{n+1}
$$

Replace $alpha$ by $displaystyle;frac{k}{y_n}$ for generic positive integer $k$, we obtain

$$left( frac{x_n}{x_n + k} right)^n = left( frac{n y_n}{n y_n + k} right)^n > left(frac{(n+1)y_n}{(n+1)y_n + k}right)^{n+1}$$
Summing over $k$ and using definition of $x_n$, we find

$$Phi_{n+1}(x_{n+1}) = 1 = Phi_n(x_n) > Phi_{n+1}((n+1)y_n)$$

Since $Phi_{n+1}$ is increasing, we obtain $x_{n+1} > (n+1)y_n iff y_{n+1} > y_n$.
This means $y_n$ is an increasing sequence.

We are going to show $y_n$ is bounded from above by $frac32$
(see update below for a more elementary and better upper bound).
For simplicity, let us abberivate $x_n$ and $y_n$ as $x$ and $y$. By their definition, we have

$$frac{2}{x^n} = sum_{k=0}^infty frac{1}{(x+k)^n}$$

By Abel-Plana formula, we can transform the sum on RHS to integrals. The end result is

$$begin{align}frac{3}{2x^n} &= int_0^infty frac{dk}{(x+k)^n} +
i int_0^infty frac{(x+it)^{-n} - (x-it)^{-n}}{e^{2pi t} - 1} dt\
&=frac{1}{(n-1)x^{n-1}}
+ frac{1}{x^{n-1}}int_0^infty frac{(1+is)^{-n} - (1-is)^{-n}}{e^{2pi x s}-1} ds
end{align}
$$
Multiply both sides by $nx^{n-1}$ and replace $s$ by $s/n$, we obtain

$$begin{align}frac{3}{2y} - frac{n}{n-1} &=
i int_0^infty frac{(1 + ifrac{s}{n})^{-n} - (1-ifrac{s}{n})^{-n}}{e^{2pi ys} - 1} ds\
&= 2int_0^infty frac{sinleft(ntan^{-1}left(frac{s}{n}right)right)}{left(1 + frac{t^2}{n^2}right)^{n/2}} frac{ds}{e^{2pi ys}-1}tag{*1}
end{align}
$$
For the integral on RHS, if we want its integrand to be negative, we need

$$ntan^{-1}left(frac{s}{n}right) > pi
implies frac{s}{n} > tanleft(frac{pi}{n}right) implies s > pi$$

By the time $s$ reaches $pi$, the factor $frac{1}{e^{2pi ys} - 1}$ already drops to very small. Numerically, we know $y_4 > 1$, so for $n ge 4$ and $s ge pi$, we have

$$frac{1}{e^{2pi ys} - 1} le frac{1}{e^{2pi^2} - 1} approx 2.675 times 10^{-9}$$

This implies the integral is positive. For $n ge 4$, we can deduce

$$frac{3}{2y} ge frac{n}{n-1} implies y_n le frac32left(1 - frac1nright) < frac32$$

Since $y_n$ is increasing and bounded from above by $frac32$, limit
$y_infty stackrel{def}{=} lim_{ntoinfty} y_n$ exists and $le frac32$.

For fixed $y > 0$, with help of DCT, one can show the last integral of $(*1)$
converges.

This suggests $y_infty$ is a root of following equation near $frac32$

$$frac{3}{2y} = 1 + 2int_0^infty frac{sin(s)}{e^{2pi ys} - 1} ds$$

According to DLMF,
$$int_0^infty e^{-x} frac{sin(ax)}{sinh x} dx = frac{pi}{2}cothleft(frac{pi a}{2}right) - frac1aquadtext{ for }quad a ne 0$$

We can transform our equation to

$$frac{3}{2y} = 1 + 2left[frac{1}{4y}cothleft(frac{1}{2y}right) - frac12right]
iff cothleft(frac{1}{2y}right) = 3$$

This leads to $displaystyle;y_infty = frac{1}{log 2}$.

This is consistent with the finding of another answer (currently deleted):

If $L_infty = lim_{ntoinfty}frac{n}{x_n}$ exists, then $L_infty = log 2$.

To summarize, the limit $displaystyle;frac{x_n}{n}$ exists and should equal to $displaystyle;frac{1}{log 2}$.

Update

It turns out there is a more elementary proof that $y_n$ is bounded from above by the optimal bound $displaystyle;frac{1}{log 2}$.

Recall for any $alpha > 0$. we have $1 + alpha < e^alpha$. Substitute
$alpha$ by $frac{k}{n}log 2$ for $n ge 2$ and $k ge 1$, we get

$$frac{n}{n + klog 2} = frac{1}{1 + frac{k}{n}log 2} > e^{-frac{k}{n}log 2} = 2^{-frac{k}{n}}$$

This leads to

$$Phi_nleft(frac{n}{log 2}right)
= sum_{k=1}^infty left(frac{n}{n + log 2 k}right)^n
> sum_{k=1}^infty 2^{-k}
= 1 = Phi_n(x_n)
$$
Since $Phi_n(x)$ is increasing, this means
$displaystyle;frac{n}{log 2} > x_n$ and $y_n$ is bounded from above by $displaystyle;frac{1}{log 2}$.

Consider the functions
$$f_n(x):=sum_{k=1}^inftyleft(frac{x}{x+k}right)^n.$$
(The series should converge for every fixed $xgeq 0$ and $ngeq 2$.)
Then the values $x_n$ are the solutions of
$$f_n(x)=1.$$
We have that $f_n(0)=0$ and because of
$$f_n'(x)=sum_{k=1}^{infty}nleft(frac{x}{x+k}right)^{n-1}frac{k}{(x+k)^2},$$
we have $f'_n(x)>0$ for $x>0$.
Moreover
$$f_n(3n)=sum_{k=1}^{infty}left(frac{3n}{3n+k}right)^ngeq3left(frac{3n}{3n+3}right)^n=3left(1+frac{1}{n}right)^{-n}.$$
Since $lim_{ntoinfty}(1+frac{1}{n})^n=e$ we have $$lim_{ntoinfty}f_n(3n)geqfrac{3}{e}>1$$ and there exists $Ninmathbb N$, such that
$$f_n(3n)>1$$
for all $ngeq N$.

Thus, for large enough $n$ we have $x_nin(0,3n)$ and
$$0leqlim_{ntoinfty}frac{x_n}{n}leq 3$$

Below is my thought of proving $limlimits_{nto infty}frac{x_n}{n}=frac{1}{ln 2}$.

For any $lambda >0$,
begin{align*}
Phi_n(lambda n)=sum_{k=1}^infty left( frac{lambda n}{lambda n+k}right)^n
end{align*}
We denote $a_{n,k}=left( frac{lambda n}{lambda n+k}right)^n$,it's easy to verify that $a_{n,k}$ is decreasing for $n$,and
begin{align*}
lim_{nto infty}a_{n,k}=e^{-k/lambda}triangleq b_k
end{align*}
We notice that $sum_{k=1}^infty b_k=sum_{k=1}^infty e^{-k/lambda}=frac{1}{e^{1/lambda}-1}$,$a_{n,k}<a_{2,k}$,$ngeq 2$,$sum a_{2,k}$is convergent．Meanwhile ,we can verify the following proposition(A similar to Lebesgue's dominated convergent theorem)

Suppose${a_{n,k}}$is a positive binary index sequence,and for all $kin mathbb{N}_+$we have
$a_{n,k}to b_k$,$ntoinfty$,besides $|a_{n,k}|<a_k$, $sum_{k=1}^infty a_k$ is convergent.Then
begin{align*}
lim_{nto infty}sum_{k=1}^infty a_{n,k}=sum_{k=1}^infty b_k
end{align*}

So thanks to the above proposition can see
begin{align*}
lim_{nto infty}Phi_n(lambda n)=sum_{k=1}^infty e^{-k/lambda}=frac{1}{e^{1/lambda}-1}
end{align*}

Specially，we take $lambda=frac{1}{ln 2}$,then $lim_{nto infty}Phi_nleft(frac{ n}{ln 2}right)=1=Phi_n(x_n)$．Thus for all $s>frac{1}{ln 2}$,since
begin{align*}
lim_{nto infty }Phi_n(s n)=frac{1}{e^{1/s}-1}>1=lim_{nto infty}Phi_n(x_n)
end{align*}
we see that there exists $N$，such that for all$ n>N$,
begin{align*}
Phi_n(s n)>Phi_n(x_n)Rightarrow sn>x_n,forall n>N
end{align*}
This implies $A=limlimits_{nto infty }y_nleqslant s$,thus $Aleqslant frac{1}{ln 2}$．Similarly we can prove $Ageqslant frac{1}{ln 2}$,and finally we get $A=frac{1}{ln 2}$.

We can rewrite $$x^{-n} = sum_{k=1}^infty (x+k)^{-n}$$

as

$$1= sum_{k=1}^infty e^{- nln (1+ k/x_n)}.$$

Now

$ln (1+k/x_n) le k/x_n$, therefore

$$1 le sum_{k=1}^infty e^{-frac{n}{x_n}k} = frac{1}{e^{n/x_n}-1}.$$

From this it follows that

$$ (*) quad n /x_n ge ln 2.$$

Suppose now that $limsup_{ntoinfty} n/x_n=M>c$. Then for all $n$ large, we have $n/x_n>c$ and

begin{align*} 1 &= sum_{k=1}^infty e^{-n ln (1+frac{k}{n} times frac{n}{x_n})}\
& le sum_{k=1}^infty e^{-n ln (1+ frac{k}{n} c)}\
& = sum_{k=1}^infty (1+frac{k}{n}c)^{-n} \
& to sum_{k=1}^infty e^{-kc}=frac{1}{e^c-1}.
end{align*}
by dominated convergence (note: $(1+frac{k}{n}c)^{-n} le (1+frac{kc}{2})^{-2}$).

Thus, $e^c-1 le 1$, or $c le ln 2$. It follows that

$$(**) quad limsup n/x_n le ln 2.$$

Now $(*)$ and $(**)$ give

$$lim_{ntoinfty} frac{x_n}{n} = sup_{n} frac{x_n}{n} = frac{1}{ln 2}.$$