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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 $$\Big|1-\frac{x_n}{n-1}(1+\frac{1}{x_n})^{1-n}\Big|<\frac{1}{(1+\frac1{x_n})^n}$$

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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

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

This leads to

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

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  • $\begingroup$ 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 $\frac1{\log 2}$. $\endgroup$ – mbfkk Nov 8 '18 at 11:28
  • $\begingroup$ @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. $\endgroup$ – achille hui Nov 8 '18 at 11:44
  • $\begingroup$ I have got a proof that $y_\infty=\frac{1}{\ln 2}$,see the third floor. $\endgroup$ – mbfkk Nov 9 '18 at 8:44
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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 $$0\leq\lim_{n\to\infty}\frac{x_n}{n}\leq 3$$

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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 $\sum_{k=1}^\infty b_k=\sum_{k=1}^\infty e^{-k/\lambda}=\frac{1}{e^{1/\lambda}-1}$,$a_{n,k}<a_{2,k}$,$n\geq 2$,$\sum a_{2,k}$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 $|a_{n,k}|<a_k$, $\sum_{k=1}^\infty a_k$ 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 $A=\frac{1}{\ln 2}$.

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  • $\begingroup$ (+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 $\frac{1}{\log 2}$. $\endgroup$ – achille hui Nov 9 '18 at 11:00
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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

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

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