# How to show the existence of the limit $\lim_{n\to \infty}\frac{x_n}{n}$ if $x_n$ satisfy $x^{-n}=\sum_{k=1}^\infty (x+k)^{-n}$?

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

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 $$x_n\ge x_{n+1}$$,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 $$\sum_{k=1}^\infty\frac{1}{(1+\frac{k}{x_n})^n}>\sum_{k=1}^\infty\frac{1}{(1+\frac{k}{x_{n+1}})^{n+1}}$$ 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 $$\lim_{n\to \infty}\frac{x_n}{n}$$.What I can see by the area's principle is

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

$$\left(1 + \frac{\alpha}{n}\right)^{n/n+1} > \frac1{n+1} \left[n\left(1 + \frac{\alpha}{n}\right) + 1 \right] = 1 + \frac{\alpha}{n+1}$$ 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^{2\pi 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^{2\pi 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 + i\frac{s}{n})^{-n} - (1-i\frac{s}{n})^{-n}}{e^{2\pi ys} - 1} ds\\ &= 2\int_0^\infty \frac{\sin\left(n\tan^{-1}\left(\frac{s}{n}\right)\right)}{\left(1 + \frac{t^2}{n^2}\right)^{n/2}} \frac{ds}{e^{2\pi ys}-1}\tag{*1} \end{align} For the integral on RHS, if we want its integrand to be negative, we need

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

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

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

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

$$\frac{3}{2y} \ge \frac{n}{n-1} \implies y_n \le \frac32\left(1 - \frac1n\right) < \frac32$$

Since $$y_n$$ is increasing and bounded from above by $$\frac32$$, limit $$y_\infty \stackrel{def}{=} \lim_{n\to\infty} y_n$$ 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 $$\frac32$$

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

According to DLMF, $$\int_0^\infty e^{-x} \frac{\sin(ax)}{\sinh x} dx = \frac{\pi}{2}\coth\left(\frac{\pi a}{2}\right) - \frac1a\quad\text{ for }\quad a \ne 0$$

We can transform our equation to

$$\frac{3}{2y} = 1 + 2\left[\frac{1}{4y}\coth\left(\frac{1}{2y}\right) - \frac12\right] \iff \coth\left(\frac{1}{2y}\right) = 3$$

This leads to $$\displaystyle\;y_\infty = \frac{1}{\log 2}$$.

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

If $$L_\infty = \lim_{n\to\infty}\frac{n}{x_n}$$ 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

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

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

• Nice done! Thanks for your reply.By the way,how can we prove that the limit is $\frac1{\log 2}$?,i.e. it's no less than $\frac1{\log 2}$. – mbfkk Nov 8 '18 at 11:28
• @mbfkk I don't have a 'rigorous' proof that $y_\infty = \frac{1}{\log 2}$, otherwise I would include that in my answer. I've already tried a few tricks but none of them work. – achille hui Nov 8 '18 at 11:44
• I have got a proof that $y_\infty=\frac{1}{\ln 2}$,see the third floor. – mbfkk Nov 9 '18 at 8:44

Consider the functions $$f_n(x):=\sum_{k=1}^\infty\left(\frac{x}{x+k}\right)^n.$$ (The series should converge for every fixed $$x\geq 0$$ and $$n\geq 2$$.) Then the values $$x_n$$ are the solutions of $$f_n(x)=1.$$ We have that $$f_n(0)=0$$ and because of $$f_n'(x)=\sum_{k=1}^{\infty}n\left(\frac{x}{x+k}\right)^{n-1}\frac{k}{(x+k)^2},$$ we have $$f'_n(x)>0$$ for $$x>0$$. Moreover $$f_n(3n)=\sum_{k=1}^{\infty}\left(\frac{3n}{3n+k}\right)^n\geq3\left(\frac{3n}{3n+3}\right)^n=3\left(1+\frac{1}{n}\right)^{-n}.$$ Since $$\lim_{n\to\infty}(1+\frac{1}{n})^n=e$$ we have $$\lim_{n\to\infty}f_n(3n)\geq\frac{3}{e}>1$$ and there exists $$N\in\mathbb N$$, such that $$f_n(3n)>1$$ for all $$n\geq N$$.

Thus, for large enough $$n$$ we have $$x_n\in(0,3n)$$ and $$0\leq\lim_{n\to\infty}\frac{x_n}{n}\leq 3$$

Below is my thought of proving $$\lim\limits_{n\to \infty}\frac{x_n}{n}=\frac{1}{\ln 2}$$.

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_{n\to \infty}a_{n,k}=e^{-k/\lambda}\triangleq b_k \end{align*} We notice that $$\sum_{k=1}^\infty b_k=\sum_{k=1}^\infty e^{-k/\lambda}=\frac{1}{e^{1/\lambda}-1}$$,$$a_{n,k},$$n\geq 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 $$k\in \mathbb{N}_+$$we have $$a_{n,k}\to b_k$$,$$n\to\infty$$,besides $$|a_{n,k}|, $$\sum_{k=1}^\infty a_k$$ is convergent.Then \begin{align*} \lim_{n\to \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_{n\to \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 $$\lim_{n\to \infty}\Phi_n\left(\frac{ n}{\ln 2}\right)=1=\Phi_n(x_n)$$．Thus for all $$s>\frac{1}{\ln 2}$$,since \begin{align*} \lim_{n\to \infty }\Phi_n(s n)=\frac{1}{e^{1/s}-1}>1=\lim_{n\to \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 $$A=\lim\limits_{n\to \infty }y_n\leqslant s$$,thus $$A\leqslant \frac{1}{\ln 2}$$．Similarly we can prove $$A\geqslant \frac{1}{\ln 2}$$,and finally we get $$A=\frac{1}{\ln 2}$$.

• (+1) good job, this settles the limit $A$ is $\frac{1}{\log 2}$. In fact, we no longer need to assume $A$ exists to get its value. For any $s > \frac{1}{\log 2}$, $y_n \le s$ for sufficiently large $n$ implies $\limsup\limits_{n\to\infty} y_n \le s$. This in turn implies $\limsup\limits_n y_n \le \inf s = \frac{1}{\log 2}$. Similarly, we have $\frac{1}{\log 2} \le \liminf\limits_{n\to\infty} y_n$. Sim limsup = liminf, limit exists and equal to $\frac{1}{\log 2}$. – achille hui Nov 9 '18 at 11:00

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

as

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

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 $$\limsup_{n\to\infty} n/x_n=M>c$$. 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

$$\lim_{n\to\infty} \frac{x_n}{n} = \sup_{n} \frac{x_n}{n} = \frac{1}{\ln 2}.$$