Equivalent of a recursively defined sequence Let $u_n$ the real sequence defined by $u_0=1$ and $$u_n=\frac{1}{n}\sum_{k=0}^{n-1} \frac{u_k}{n-k}.$$
The goal is to show that there exists $C>0$ such that $u_n \sim C/n^2$. 
I am only able to prove that $u_n =O(\log(n)/n^2)$. I tried to use an approached based on Cauchy product of power series but it failed. 
 A: The "Cauchy product" idea is good to start with. Introducing $u(z)=\sum\limits_{n=0}^\infty u_n z^n$, we have $$zu'(z)=\sum_{n=1}^\infty nu_n z^n=\sum_{n=1}^\infty\sum_{k=0}^{n-1}u_k z^k\frac{z^{n-k}}{n-k}=-u(z)\ln(1-z),$$ that is, $\color{blue}{u(z)=e^{\mathrm{Li}_2(z)}}$ where $\mathrm{Li}_2(z)=-\displaystyle\int_0^z\frac{\ln(1-w)}{w}~dw=\sum_{n=1}^\infty\frac{z^n}{n^2}$ is the dilogarithm.
Intermezzo. Here, $\sum\limits_{n=1}^\infty n^2 u_n z^n=z\big(zu'(z)\big)'=\left(\frac{z}{1-z}+\ln^2(1-z)\right)u(z)$. Since $$\lim_{n\to\infty}a_n=a\text{ exists}\implies a=\lim_{x\to 1^-}(1-x)\sum_{n=1}^\infty a_n x^n,$$ we know that if $C=\lim\limits_{n\to\infty} n^2 u_n$ exists, then $\color{blue}{C=e^{\pi^2/6}}$. But that "if" seems hard to get rid of...
I'm going the complex-analytic way below. We have $u_n=\frac{1}{2\pi i}\int_L\frac{u(z)}{z^{n+1}}~dz$, where $L$ is any (simple, positively oriented, lying in the domain of analyticity of the integrand) contour encircling $0$. Now $\mathrm{Li}_2(z)$ has an analytic extension onto the entire complex plane $(z)$ cut along $[1,+\infty)$, and we have the identity $$\mathrm{Li}_2(z)+\mathrm{Li}_2(1/z)=-\frac{\pi^2}{6}-\frac{\ln^2(-z)}{2}\qquad(z\notin\mathbb{R}_+)$$ Now take $R>1$ and let $L$ be the circle $|z|=R$ with a notch around the cut. The above identity gives $|u(z)|=\mathcal{O}(e^{-(\ln R)^2/2})$ uniformly on $|z|=R$, and the integral over it tends to $0$ with $R\to\infty$. Hence we can let $L$ be a Hankel-type contour encircling $[1,+\infty)$; substituting $z=1+w/n$, we get $$u_n=\frac{1}{2n\pi i}\int_\lambda\frac{u(1+w/n)}{(1+w/n)^{n+1}}~dw,$$ where $\lambda$ is a Hankel-type contour encircling $[0,+\infty)$. Using the above identity again, we arrive at $$n^2 u_n=\int_0^\infty\frac{\exp\{\frac{\pi^2}{3}-\mathrm{Li}_2\left(\frac{n}{n+x}\right)-\frac{1}{2}\ln^2\left(1+\frac{x}{n}\right)\}}{\left(1+\frac{x}{n}\right)^{n+1}}\frac{n}{\pi}\sin\left\{\pi\ln\left(1+\frac{x}{n}\right)\right\}~dx,$$ and the dominated convergence theorem is applicable here, giving $$\lim_{n\to\infty}n^2 u_n=\int_0^\infty\frac{e^{\pi^2/6}}{e^x}x~dx=e^{\pi^2/6}.$$
