In a much larger problem, I need to solve for $x_n$ the equation $$\sum_{i=0}^n (n+i)^{x_n}=(2n+1)^{x_n}$$ where $n$ can be large.

From a numerical point of view, it does not make any practical difficulty solving instead $$\log\left(\sum_{i=0}^n (n+i)^{x_n} \right)={x_n}\log(2n+1)$$ which is "almost" finding the intersection of two straight lines.

My first surprise, working with small values of $n$, was to notice that the solution is almost linearly dependent on $n$.

Working with large numbers, what I noticed is the following $$\left( \begin{array}{cc} n & \frac {x_n} n \\ 10^1 & 1.351611135 \\ 10^2 & 1.382828492 \\ 10^3 & 1.385947786 \\ 10^4 & 1.386259704 \end{array} \right)$$ and the ratio "seems" to converge to something.

So, my questions :

  • Could we prove the linear dependency of $x_n$ to $n$ ?
  • Could we prove that $\frac {x_n} n$ tends to a limit ?
  • If there is a limit, what could be this number ? In other words, would it be possible to establish the asymptotics of $x_n$ ?

Edit after answers

After the beautiful answers I received from Professor Vector and Raymond Manzoni (who made the problem more general), I solved, for $n=10^4$, the equation$$\sum_{i=0}^n (n+i)^{x_n}=(2n+a)^{x_n}$$ In the table below are reproduced the results for the computed ratio $\frac {x_n} n$ and the value computed for $y_a=-2\log(b)$ where $b$ is the positive solution of $1-b=b^a$. $$\left( \begin{array}{cc} a & \frac {x_n} n & y_a\\ 1 & 1.38626 &1.38629 \\ 2 & 0.96240 &0.96242 \\ 3 & 0.76448 &0.76449 \\ 4 & 0.64457 &0.64457 \\ 5 & 0.56240 &0.56240 \\ 6 & 0.50184 &0.50183 \\ 7 & 0.45496 &0.45495 \\ 8 & 0.41739 &0.41737 \\ 9 & 0.38646 &0.38644 \end{array} \right)$$

Concerning the case where $a=1$, it is interesting to compare the results I gave with the asymptotics Raymond Manzoni gave in a comment. $$\frac {x_n} n=2\log(2)-\frac{\log(2)}{2n}+O\left(\frac 1 {n^2} \right)$$ $$\left( \begin{array}{cc} n & \frac {x_n} n &2\log(2)-\frac{\log(2)}{2n}\\ 10^1 & 1.351611135 &1.351637002\\ 10^2 & 1.382828492 &1.382828625\\ 10^3 & 1.385947786 &1.385947788\\ 10^4 & 1.386259704 &1.386259704 \end{array} \right)$$ I do not think that I could add any comment !

  • 3
    $\begingroup$ Of course, the closeness to $2\ln2=1.38629436111\ldots$ could be a coincidence, but I guess it's more. $\endgroup$ – Professor Vector Aug 1 '17 at 9:34
  • $\begingroup$ @ProfessorVector. I noticed that but I preferd not to mention it. It could be funny, isn't it ? $\endgroup$ – Claude Leibovici Aug 1 '17 at 9:37
  • $\begingroup$ I think I can prove that limit, but not at work. $\endgroup$ – Professor Vector Aug 1 '17 at 10:08
  • $\begingroup$ @ProfessorVector. I shall wait, then ! Thanks. $\endgroup$ – Claude Leibovici Aug 1 '17 at 10:10
  • 2
    $\begingroup$ Yay! $2\ln\phi=0.962423650119\ldots$ ($\phi$ being Golden Ratio) $\endgroup$ – Professor Vector Aug 1 '17 at 10:30

Let's investigate the equation $$\sum_{i=0}^n (n+i)^{x_n}=(2n+1)^{x_n}:$$ As $i$ goes from $0$ to $n$, $n+i$ goes from $n$ to $2n$. If we reorder the sum upside down, and divide both sides by $(2n+1)^{x_n}$, we get $$\sum_{k=1}^{n+1}\left(\frac{2n+1-k}{2n+1}\right)^{x_n}=\sum_{k=1}^{n+1}\left(1-\frac{k}{2n+1}\right)^{x_n}=1.$$ Using the elementary inequality $$1-\frac{k}{2n+1}\le\left(1-\frac1{2n+1}\right)^k$$ and setting $$\lambda_n=\left(1-\frac1{2n+1}\right)^{x_n},$$ the equation gives $$1\le\sum_{k=1}^{n+1}(\lambda_n)^k<\sum_{k=1}^\infty(\lambda_n)^k=\frac{\lambda_n}{1-\lambda_n}.$$ So $\lambda_n>1/2,$ meaning $$x_n<\frac{\ln 2}{\ln\frac1{1-1/(2n+1)}}<(2n+1)\ln 2,\tag{1}$$ since $\ln\frac1{1-x}>x$ for $1>x>0.$
The lower bound for $x_n$ is somewhat more technical:
We have $$1-\frac{k}{2n+1}=\frac1{1+\frac{k}{2n+1-k}}\ge\left(\frac1{1+\frac{1}{2n+1-k}}\right)^k\ge\left(\frac1{1+\frac{1}{2n+1-\sqrt{n}}}\right)^k$$ for $k\le\sqrt{n}.$ Setting $$\mu_n=\left(\frac1{1+\frac{1}{2n+1-\sqrt{n}}}\right)^{x_n},$$ we obtain the estimate $$1\ge\sum_{k=1}^{\lfloor\sqrt{n}\rfloor}\left(1-\frac{k}{2n+1}\right)^{x_n}\ge\sum_{k=1}^{\lfloor\sqrt{n}\rfloor}(\mu_n)^k=\frac{\mu_n}{1-\mu_n}\left(1-(\mu_n)^{\lfloor\sqrt{n}\rfloor}\right).$$ Clearly, this means $\mu_n\le1/(2-\epsilon_n),$ where $\epsilon_n=O(c^\sqrt{n})$ with some $c<1.$ This means $$x_n\ge\frac{\ln(2-\epsilon_n)}{\ln\left(1+\frac{1}{2n+1-\sqrt{n}}\right)}>(2n+1-\sqrt{n})\ln(2-\epsilon_n),\tag{2}$$ since $\ln(1+x)<x$ for $x>-1.$
(1) and (2) together imply $$\left(2+\frac1n-\frac1{\sqrt{n}}\right)\ln(2-\epsilon_n)<\frac{x_n}{n}<\left(2+\frac1n\right)\ln2,$$ and thus, $$\lim_{n\rightarrow\infty}\frac{x_n}{n}=2\ln2.$$
In the case of the equation $$\sum_{i=0}^n (n+i)^{x_n}=(2n+2)^{x_n},$$ we get $$\sum_{k=2}^{n+2}\left(1-\frac{k}{2n+2}\right)^{x_n}=1,$$ so the equation for $\lambda=\lim_{n\rightarrow\infty}\lambda_n$ becomes not $\lambda/(1-\lambda)=1,$ but $\lambda^2/(1-\lambda)=1,$ i.e. $\lambda=1/\phi.$

  • $\begingroup$ This is really a beautiful and elegant answer to the problem. Thank you very much ! $\endgroup$ – Claude Leibovici Aug 2 '17 at 4:12
  • $\begingroup$ One direction is a bit technical and not so pretty, but that's inevitable, because your sum is dominated by far less than $n$ largest terms $(1+\epsilon)\ln n,$ I'd say. I used $\sqrt{n}$ only as a convenient compromise. $\endgroup$ – Professor Vector Aug 2 '17 at 4:15

Let's use Euler–Maclaurin to approximate your sum for large $n$ ($p$ is a fixed integer) : $$\sum_{i=0}^n f(i) = \int^n_0 f(t)\,dt + \frac{f(n) + f(0)}{2} + \sum_{k=1}^{\lfloor p/2\rfloor} \frac{B_{2k}}{(2k)!} (f^{(2k - 1)}(n) - f^{(2k - 1)}(0)) + R$$ For $\,f(t):=(n+t)^x\;$ this becomes : \begin{align} \sum_{i=0}^n (n+i)^x &= \int^n_0 (n+t)^x\,dt + \frac{(2n)^x + (n)^x}{2} + \sum_{k=1}^{\lfloor p/2\rfloor} \frac{B_{2k}}{(2k)!} (f^{(2k - 1)}(n) - f^{(2k - 1)}(0)) + R\\ &=\frac{\left(2^{x+1}-1\right)n^{x+1}}{x+1}+ \frac{(2^x+1)n^x}{2} + \sum_{k=1}^{\lfloor p/2\rfloor} \frac{B_{2k}}{(2k)!} (f^{(2k - 1)}(n) - f^{(2k - 1)}(0)) + R\\ \end{align}

Let's first neglect the Bernoulli terms and remainder then we want $x$ solution of :

$$\frac{\left(2^{x+1}-1\right)n^{x+1}}{x+1}+ \frac{(2^x+1)n^x}{2}\approx(2n+a)^{x}$$ for $n\gg 1\,$ and $\,x:=ny\;$ we may neglect the $\pm 1$ terms to get : $$\frac{(2n)^{ny+1}}{ny}+\frac{(2n)^{ny}}{2}\approx(2n)^{ny}\left(1+\frac a{2n}\right)^{2n\,y/2}$$ simplification and limit as $n\to \infty$ at the right leaves : $$\frac{2}{y}+\frac{1}{2}\approx e^{\,ay/2}$$ admitting the approximate solution $\,y\approx 1.358\,$ for $a=1$.

Incorporating the two first Bernoulli corrections we get instead : $$\frac 12+\frac 2y+\frac y{24}-\frac {y^3}{5760}\approx e^{\,ay/2}$$ where $\dfrac 1{24}=\dfrac {B_{2}}{2^1 2!}$ and $\dfrac {-1}{5760}=\dfrac {B_{4}}{2^3 4!}\;$ suggesting us to add all the Bernoulli terms to obtain the usual generating function $\;\displaystyle \frac 1{1-e^{-y/2}}\;$ at the left. As $n\to\infty$ the problem becomes to solve : $$1-e^{-y/2}=e^{-ay/2}$$ i.e. obtain the positive solution of $\,1-b=b^a\;$ and conclude with $\;y=2\ln(1/b)$.

For $a=1\,$ this returns the wished $\,y=2\ln(2)$ while for $a=2\,$ you'll obtain $y=2\ln(\phi)$ ($\phi$ the golden ratio) : both solutions proposed by Professor Vector (+1) !

ADDITIONS: I will further conjecture following expansion for $a=1$ : $$\frac {x_n}n=\ln(2)\left(2-\frac 1{2n}+\frac {-\frac{25}{24}+\frac 32\ln(2)}{n^2}+\frac{-\frac {169}{96}+\frac{55}8\ln(2)-\frac{25}4\ln(2)^2}{n^3}\right)+O\left(\frac 1{n^4}\right)$$

Claude Leibovici suggested (privately) to replace $n$ by a fixed $a$ in the previous problem that is to solve $\;\displaystyle\sum_{i=0}^n (a+i)^x= (a+n+k)^{x}$.
Let's use the same method with $\,f(t):=(a+t)^x\;$ : $$\sum_{i=0}^n (a+i)^x \sim \frac{(a+n)^{x+1}-a^{x+1}}{x+1}+ \frac{(a+n)^x + a^x}{2} + \sum_{k=1}^{\lfloor p/2\rfloor} \frac{B_{2k}}{(2k)!} (f^{(2k - 1)}(n) - f^{(2k - 1)}(0))$$

without the Bernoulli terms we would have :

$$\frac{(a+n)^{x+1}-a^{x+1}}{x+1}+ \frac{(a+n)^x + a^x}{2} \approx(a+n+k)^{x}$$

for $\,n\gg 1\,$ and $\,x=ny\,$ we may neglect the powers of $a$ alone and the $+1$ : $$\frac{(a+n)^{ny+1}}{ny}+\frac{(a+n)^{ny}}{2}\approx(a+n)^{ny}\left(1+\frac k{a+n}\right)^{(a+n)\,y-ay}$$

After division by $(a+n)^{ny}$ and for $n \gg 1$ : $$\frac{a+n}{ny}+\frac 12\approx e^{ky}\left(1+\frac k{a+n}\right)^{-ay}$$

or as $ n\to\infty$ : $$\frac 1y+\frac{1}{2}\approx e^{\,ky}$$ with a solution $y$ near to $\ln(2)$ for $k=1$.

Adding all the Bernoulli terms gives at the left $\dfrac 1{1-\exp(-y)}\;$ so that we must solve : $1-\exp(-y)=\exp(-ky)$ or, again, $1-b=b^k$ followed this time by $y = \ln(1/b)$.

I'll conclude with this (conjectured) expansion for $k=1$ : $$\frac {x_n}n = \ln(2)\left(1+\frac{a-1/2}n +\frac {-\frac{25}{12}+3\ln(2)}{n^2}\right)+ O\left(\frac 1{n^3}\right)$$

  • $\begingroup$ Hi Raymond ! Long time no speak. This is really a beautiful and elegant answer to the problem. Thank you very much ! $\endgroup$ – Claude Leibovici Aug 2 '17 at 4:13
  • $\begingroup$ Glad you liked this @Claude (fine question and edit btw !). Expansions in negative powers of $n$ are of interest too getting $\;\displaystyle x_n=2\ln(2)-\frac {\ln(2)}{2\,n}+O\left(\frac 1{n^2}\right)\;$ and more if needed. Cheers, $\endgroup$ – Raymond Manzoni Aug 2 '17 at 6:36
  • $\begingroup$ My method can't give the asymptotics, it's too elementary for that. I did compute the limit for $a=3$, though, it's 0.76449017168007128265871699836971. $\endgroup$ – Professor Vector Aug 2 '17 at 7:02
  • $\begingroup$ @Professor Vector: Corresponding to alpha's answer : $$2\ln\left(2+2^{2/3}\sqrt[3]{29-3\sqrt{93}}+2^{2/3}\sqrt[3]{29+3\sqrt{93}}\right)-2\ln(6)$$. Answer of different kinds are always appreciated! Cheers, $\endgroup$ – Raymond Manzoni Aug 2 '17 at 7:23
  • $\begingroup$ @Claude Leibovici: I added two more detailed expansions that could interest you! Cheers, $\endgroup$ – Raymond Manzoni Aug 4 '17 at 0:26

Using Bernoulli's Inequality, $$ \begin{align} \frac{\left(1-\frac{k+1}{2n+3}\right)^{n+\frac32}}{\left(1-\frac{k+1}{2n+1}\right)^{n+\frac12}} &=\left(1-\frac{k+1}{2n+3}\right)\left(\frac{2n+2-k}{2n+3}\frac{2n+1}{2n-k}\right)^{n+\frac12}\\ &=\left(1-\frac{k+1}{2n+3}\right)\left(1+\frac{2k+2}{(2n+3)(2n-k)}\right)^{n+\frac12}\\[9pt] &\ge\left(1-\frac{k+1}{2n+3}\right)\left(1+\frac{(2n+1)(k+1)}{(2n+3)(2n-k)}\right)\\[9pt] &=1+\frac{2(k+1)^2}{(2n-k)(2n+3)^2}\\[12pt] &\ge1 \end{align} $$ That is, $\left(1-\frac{k+1}{2n+1}\right)^{n+\frac12}$ is increasing in $n$.

Therefore, by Monotone Convergence, $$ \begin{align} \lim_{n\to\infty}\sum_{k=0}^n\left(\frac{n+k}{2n+1}\right)^{\left(n+\frac12\right)a} &=\lim_{n\to\infty}\sum_{k=0}^n\left(\frac{2n-k}{2n+1}\right)^{\left(n+\frac12\right)a}\\ &=\lim_{n\to\infty}\sum_{k=0}^n\left(1-\frac{k+1}{2n+1}\right)^{\left(n+\frac12\right)a}\\ &=\sum_{k=0}^\infty e^{-(k+1)a/2}\\ &=\frac{e^{-a/2}}{1-e^{-a/2}} \end{align} $$ The $a$ that gives the limit equal to $1$ is $$ a=2\log(2) $$

  • $\begingroup$ Strictly speaking, you don't show that the exact solution $x_n$ is approximately $2\ln2(n+1/2)$ (ratio of arguments $\rightarrow1$ as $n\rightarrow\infty$), but that the latter is an approximate solution (function values $\rightarrow1$). That's not necessarily the same thing, but equivalence is easy to show in this case, as the convergence is uniform in $a$ in every finite interval, due to monotony, again (Dini's theorem). $\endgroup$ – Professor Vector Aug 2 '17 at 18:57
  • $\begingroup$ Another nice solution using a quite different path. Thank you very much ! $\endgroup$ – Claude Leibovici Aug 3 '17 at 4:30

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