# Solve the sequence : $u_n = 1-(\frac{u_1}{n} + \frac{u_2}{n-1} + \ldots + \frac{u_{n-1}}{2})$

For a physical model I am trying to solve this sequence:

\begin{align*} u_1 &= 1 \\ u_2 &= 1-\left(\frac{u_1}{2}\right) \\ u_3 &= 1-\left(\frac{u_1}{3} + \frac{u_2}{2}\right) \\ u_4 &= 1-\left(\frac{u_1}{4} + \frac{u_2}{3} + \frac{u_3}{2}\right) \\ &\vdots \\ u_n &= 1-\left(\frac{u_1}{n} + \frac{u_2}{n-1} + \cdots + \frac{u_{n-1}}{2} \right) \end{align*}

I would like to have at least the asymptotic behavior of $u_n$ but i really don't know how to do.

I tried to use the integral way to solve the harmonic series, so I take $f$ such as $f(n) = u_n$ so $$f(n) = 1-\left(\frac{f(1)}{n} + \frac{f(2)}{n-1} + \cdots + \frac{f(n-1)}{2}\right)$$ I suppose $f$ is decreasing and so I write $$f(n) \sim 1- \int_1^n \frac{f(x)}{n+1-x} \,dx$$ and then $$f'(x) + f(x) = \frac{1}{x}$$ But I'm not sure this is a correct way, and I don't really know how to solve that (seems to be $\sim 1/x$ when $x \to \infty$, but I found a solution $1 + \mathrm{Ei}(x)/e^{-x}$ that seems to be divergent).

• If my calculation is correct, we have $$\sum_{n=1}^{\infty} u_n x^n = -\frac{x^2}{(1-x)\log(1-x)}.$$ In particular, $\log |u_n| = 1+o(1)$. But this is a very poor asymptote. Also, I want to point out that $\mathrm{Ei}(x)/e^x$ converges to 0 as $x\to\infty$, in view of L'hospital's rule. – Sangchul Lee Mar 11 '13 at 13:23
• Typo in the formula for $u_4$? – Eckhard Mar 11 '13 at 13:23
• And it follows that $u_n$ has zero Abel-mean and $nu_n$ has infinite Aben-mean. Thus if $u_n$ has limit, monotone decreasing for example, the it should be zero. – Sangchul Lee Mar 11 '13 at 13:28
• You have $u_n=1-\sum_{j=1}^{n-1}u_j\frac{1}{n+1-j}$. My idea would be to investigate objects like $u_{n+m}-u_n$. Of you might be able to plug the $u_j$'s into $u_n$ and obtain something which might look like "$1-\Pi_{k=1}^n\sum_k^{?}(\frac{1}{k!}+...)u_1^k$" – Nikolaj-K Mar 11 '13 at 13:38

Here is some observations: Let

$$U(x) = \sum_{n=1}^{\infty} u_n x^n.$$

From the identity $\displaystyle -\frac{\log(1-x)}{x} = \sum_{n=0}^{\infty} \frac{x^n}{n+1}$, it follows that

$$-\frac{\log(1-x)}{x} U(x) = \sum_{n=1}^{\infty} \left( \sum_{k=1}^{n} \frac{u_k}{n+1-k} \right) x^n = \sum_{n=1}^{\infty} x^n = \frac{x}{1-x}$$

and hence

$$U(x) = -\frac{x^2}{(1-x)\log(1-x)}.$$

Now let

$$-\frac{x}{\log(1-x)} = 1 - \sum_{n=1}^{\infty} a_n x^n.$$

Then it is plain to confirm that

$$u_n = 1 - \sum_{k=1}^{n-1} a_k.$$

Regarding the behavior of $(a_n)$, we claim the following assertions:

1. $a_n > 0$ for all $n = 1, 2, 3, \cdots$ and
2. $\sum a_n = 1$.

Once $a_n > 0$ is established, then the second assertion immediately follows. Indeed, the sum $\sum a_n$ tends to some limit in $[0, \infty]$. Then a simple version of Abelian theorems shows that

$$\sum_{n=1}^{\infty} a_n = \lim_{x\to 1^-} \sum_{n=1}^{\infty} a_n x^n = \lim_{x\to 1^-} \left( 1 + \frac{x}{\log (1-x)} \right) = 1.$$

Thus it remains to prove the first assertion. Let

$$f(x) = \frac{x-1}{\log x}.$$

Then it is easy to check that $f(x) \geq 0$ on $[0, \infty)$ and

$$a_n = \frac{(-1)^{n-1}}{n!} f^{(n)}(1).$$

This shows that it suffices to check the condition $(-1)^{n-1} f^{(n)}(x) > 0$ for $x > 0$ and $n = 1, 2, 3, \cdots$. In other words, it is sufficient to show that $f$ is a Bernstein function. This follows once we show that

$$f(s) = \int_{0}^{\infty} \left( -\int_{0}^{1}\frac{t^{u-2}}{(u-2)!} \, du \right) \left( 1 - e^{-st} \right) \, dt, \tag{1}$$

since the integrand is positive inside the domain of the integration. But this follows from the following calculation:

\begin{align*} \frac{s-1}{\log s} &= s \int_{0}^{1} \frac{du}{s^u} = s \int_{0}^{1} \frac{1}{\Gamma(u)} \int_{0}^{\infty} t^{u-1} e^{-st} \, dtdu \\ &= s \int_{0}^{\infty} \left( \int_{0}^{1} \frac{t^{u-1}}{\Gamma(u)} \, du \right) e^{-st} \, dt \\ &= \left[ \left( \int_{0}^{1} \frac{t^{u-1}}{\Gamma(u)} \, du \right) \left( 1 - e^{-st} \right) \right]_{0}^{\infty} - \int_{0}^{\infty} \left( \int_{0}^{1} \frac{t^{u-1}}{\Gamma(u)} \, du \right)' \left( 1 - e^{-st} \right) \, dt. \end{align*}

Now combining the two assertions and $(1)$, we have

\begin{align*} u_n = \sum_{k=n}^{\infty} a_k &= \sum_{k=n}^{\infty} \frac{(-1)^{k-1}}{k!} f^{(k)}(1) \\ &= \sum_{k=n}^{\infty} \int_{0}^{\infty} \int_{0}^{1} \frac{1-u}{\Gamma(u)} \frac{t^{u+k-2}}{k!} e^{-t} \, du \, dt \\ &= \int_{0}^{\infty} \int_{0}^{1} \frac{1-u}{\Gamma(u)} t^{u-2} \int_{0}^{t} \frac{v^{n-1}}{(n-1)!} e^{-v} \, dv \, du \, dt \\ &= \int_{0}^{\infty} \int_{0}^{1} \frac{t^{u+n-2}}{(n-1)!\Gamma(u)}e^{-t} \, du \, dt \\ &= \int_{0}^{1} \frac{\Gamma(u+n-1)}{(n-1)!\Gamma(u)} \, du \\ &= \int_{0}^{1} \prod_{k=1}^{n-1} \left( 1 - \frac{u}{k} \right) \, du. \end{align*}

Finally, heuristic calculation then suggests that

$$\lim_{n\to\infty} u_n \log n = 1,$$

which I'm trying to prove.

$$\sum_{n=1}^{\infty} u_n x^n = -\frac{x^2}{(1-x)\log(1-x)}$$

The coefficients of $u_{n+1}$ can be obtained by a contour integral around the origin:

$$u_{n+1} = \frac{1}{2\pi i}\int_{C} \frac{dx}{x^n(x-1)\log(1-x)}\tag{*}$$ In the integrand, there is a pole at $x = 1$ and a branch cut from $\log(1-x)$ over the interval $[1,\infty)$. If we deform the contour $C$ to start from $\infty - i\epsilon$, runs along bottom of the branch cut to the pole $x = 1$, walks around the pole clockwisely and then runs along top of the branch cut back to $\infty + i\epsilon$. We can convert $(*)$ to:

$$u_{n+1} = \frac{1}{2\pi i} \int_{1}^{\infty} \frac{dt}{t(1+t)^n}\left[\frac{1}{\log(t)-\pi i} - \frac{1}{\log(t)+\pi i}\right]$$

Let $t = e^{\pi y}$, this can be simplified to: $$u_{n+1} = \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{dy}{y^2 + 1} \left(\frac{1}{e^{\pi y}+1}\right)^n$$

For large $n$, the factor $\left(\frac{1}{e^{\pi y}+1}\right)^n$ sort of behaves like a step function. When $e^{\pi y} + 1$ is close to $1$, its value is about $1$, when $e^{\pi y} + 1$ differs from $1$, its value vanishes quickly. To the lowest order, we can approximate this factor as a step function $\theta(y_0 - y)$ where $y_0$ is chosen such that the factor becomes $\frac12$, i.e:

$$\left(\frac{1}{e^{\pi y_0} + 1}\right)^n = \frac12 \implies y_0 = \frac{\log(2^{\frac{1}{n}} - 1)}{\pi} \sim \frac{\log\log(2) - \log n}{\pi}$$

$$u_{n+1} \sim \frac{1}{\pi}\int_{-\infty}^{y_0} \frac{dy}{1 + y^2} = \frac{1}{\pi} \arctan( \frac{\pi}{\log n - \log\log 2} ) \tag{**}$$

The way of choosing $y_0$ still need to be justified. This means the offset $\log\log 2$ in $(**)$ may change under more rigorous treatment. However, $(**)$ alone seems to provide a very good approximation to $u_{n+1}$. Up to what I have tested ( $20 \le n \le 300$ ), the R.H.S for $(**)$ can reproduce the value of $u_{n+1}$ with a relative error less than 1 percent.

For large $n$, we once again obtain:

$$u_{n+1} \sim \frac{1}{\pi}\frac{\pi}{\log n - \log\log 2} \sim \frac{1}{\log n}$$

Same conclusion as everyone else.

• I usually experience some sort of estranging whe I see a complex-analysis-based solution, but your solution is so appealing! I sense some flavor of analytic number theory, though I'm not sure of this feeling... – Sangchul Lee Mar 12 '13 at 12:42

Related problem: (I), (II). Here is a closed form formula for the $n$th derivative of the generating function

$$U(x)=\frac{x^2}{(1-x)\ln(1-x)}$$

at $x=0$

$$U^{(n)}(0) = (-1)^n\sum_{k=0}^{n} \left[\matrix{n\\k}\right]\frac{1+(-1)^{k+1}}{k+1},$$

where $\left[\matrix{n\\k}\right]$ are Stirling numbers of the first kind. This allows us to construct the Taylor series of the generating function at the point $x=0$ as

$$U(x)=\frac{x^2}{(1-x)\ln(1-x)}= \sum_{n=0}^{\infty}\sum_{k=0}^{n} \left[\matrix{n\\k}\right]\frac{1+(-1)^{k+1}}{k+1}\frac{(-x)^n}{n!}.$$

Now, it is clear what the closed form formula for $u_n$ is?

Once you accept sos440's integral formula, you can get the asymptotics of $u_n$ pretty easily. I think you could avoid the sinc function if necessary, but I like it.

Fix $0\leq u\leq 1$. The bounds $1-w^2\leq (1-w)\exp(w)\leq 1$ for $0\leq w\leq 1$, give
$$\prod_{k=1}^{n}\left(1-{u^2\over k^2}\right) \exp(-u H_{n}) \leq \prod_{k=1}^{n}\left(1-{u\over k}\right)\leq \exp(-u H_n),$$ where $H_n$ is the $n$th harmonic number. The infinite product expansion of the sinc function is $${\rm sinc}(\pi u)=\prod_{k=1}^{\infty}\left(1-{u^2\over k^2}\right),$$ so we have $${\rm sinc}(\pi u)\exp(-u H_n)\leq\prod_{k=1}^{n}\left(1-{u\over k}\right)\leq\exp(-u H_n).\tag1$$

Setting $u_{n+1}=\int_0^1 \prod_{k=1}^{n}\left(1-{u\over k}\right)\,du$, using the bounds (1) and a change of variables gives

$$\int_0^{H_n} {\rm sinc}(\pi w/H_{n}) \exp(-w)\,dw \leq H_n u_{n+1} \leq \int_0^{H_n} \exp(-w)\,dw,$$ and since the sinc function is bounded, continuous, and takes the value 1 at zero, using monotone convergence we conclude that $H_n u_{n+1} \to1$. In other words, $u_n\sim 1/\log(n)$.

• Oh, introducing the sinc function was my last missing ingredient! I also made a similar calculation, but it lacked a precise description for the lower bound, hence remained as a mere heuristics. +1 upvote! – Sangchul Lee Mar 12 '13 at 3:06
• @sos440 Glad to help. I was playing around with the sinc function this weekend, so it was on my mind. – user940 Mar 12 '13 at 3:15
• I suppose we have $u_n \sim 1/log(n)$ at the end. Thank you very much for this answer. – user66240 Mar 12 '13 at 10:51
• @user66240 Whoops! Thanks for pointing out my mistake. – user940 Mar 12 '13 at 12:05
• Nice and clear derivations. – Mhenni Benghorbal Mar 12 '13 at 13:40

Change the recurrence to: $$u_{n} = 1 - \sum_{0 \le k < n} \frac{u_k}{n - k + 1} \qquad u_0 = 0$$ This also gives $u_1 = 1$ and simplifies some of the following.

Define the ordinary generating function $U(z) = \sum_{n \ge 0} u_n z^n$. By properties of the generating functions, as the sum is the convolution of the secuence $\langle \frac{1}{n + 1} \rangle$ and $\langle u_n \rangle$, and: $$\sum_{n \ge 0} \frac{z^n}{n + 1} = - \frac{\ln (1 - z)}{z}$$ this gives: \begin{align*} U(z) &= \frac{1}{1 - z} + U(z) \frac{\ln(1 - z)}{z} \\ U(z) &= -\frac{z}{z^2 - z(\ln (1 - z) + 1) + \ln(1 - z)} \end{align*} Update

The denominator factors nicely: $$U(z) = - \frac{z}{(1 - z) (\ln (1 - z) - z)}$$ This has simple poles at 0 and at 1.

• Comparing with the actual values of the sequence (obtained by Mathematica, by solving the given system of equations directly) shows that your power series does not agree with the sequence. I think you unwittingly made a mistake when dealing with the Cauchy product of $u_n$ and $1/(n+1)$. – Sangchul Lee Mar 12 '13 at 3:09