Solve the sequence : $u_n = 1-(\frac{u_1}{n} + \frac{u_2}{n-1} + \ldots + \frac{u_{n-1}}{2})$
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:
- $a_n > 0$ for all $n = 1, 2, 3, \cdots$ and
- $\sum a_n = 1$.
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.
Start with the generating function
$$\sum_{n=1}^{\infty} u_n x^n = -\frac{x^2}{(1-x)\log(1-x)}$$
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:
$$\left(\frac{1}{e^{\pi y_0} + 1}\right)^n = \frac12 \implies y_0 = \frac{\log(2^{\frac{1}{n}} - 1)}{\pi} \sim \frac{\log\log(2) - \log n}{\pi} $$
This leads to an approximation:
$$u_{n+1} \sim \frac{1}{\pi}\int_{-\infty}^{y_0} \frac{dy}{1 + y^2} = \frac{1}{\pi} \arctan( \frac{\pi}{\log n - \log\log 2} ) \tag{**}$$
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.
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, $u_n\sim 1/\log(n)$.
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.