TITLE: Solve the sequence : $u_n = 1-(\frac{u_1}{n} + \frac{u_2}{n-1} + \ldots + \frac{u_{n-1}}{2})$
QUESTION [13 upvotes]: 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).

REPLY [8 votes]: 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?