A colleague of mine is interested in finding out how to show the following:
Prove that the sequence $(a_n)$ defined by
$$a_{n+1}=\frac{1}{n+1}\left(\frac{n+1}{n+2}a_0+\frac{n}{n+1}a_1+\frac{n-1}{n}a_2+\cdots+\frac{1}{2}a_n\right), \;a_0=1$$
is decreasing.
EDIT: After seeing Greg Martin's calculation, I have corrected the statement of the problem.
Let we set: $$ f(x)=\sum_{n\geq 0} a_n x^n. \tag{1}$$ Obviously $(n+1)\,a_{n+1}$ is the coefficient of $x^n$ in the Taylor series of $f'(x)$. Since: $$ g(x)=\sum_{n\geq 0}\frac{n+1}{n+2}x^n = \frac{1}{x(1-x)}+\frac{\log(1-x)}{x^2}\tag{2}$$ the recurrence relation is equivalent to the differential equation: $$ f'(x) = f(x)\cdot g(x)\tag{3} $$ from which:
$$ f(x) = \exp\left(-1-\frac{\log(1-x)}{x}\right)=\prod_{m\geq 2}\exp\left(\frac{x^{m-1}}{m}\right)\tag{4}$$
follows. Now it should be not too difficult to prove that $a_n>a_{n+1}$ as well as: $$ \lim_{n\to +\infty}a_n = \frac{1}{e}.$$ We may also notice that $f(x)$ is related with the exponential generating function for derangements, and our sequence is OEIS A055505/OEIS A055535. To prove that our sequence is decreasing, we just need to prove that any coefficient of the Taylor series of $(1-x)f(x)$, except the very first one, is negative. That follows from: $$ (1-x)\,f(x) = \exp\left(-\sum_{m\geq 1}\frac{x^m}{m}+\sum_{m\geq 1}\frac{x^m}{m+1}\right)=\exp\left(-\sum_{m\geq 1}\frac{x^m}{m(m+1)}\right),\tag{5}$$ so $(1-x)\,f(x)$ is the exponential of an analytic function with all its coefficients being negative. Now we just need to prove that: $$ \lim_{n\to +\infty}[x^n]\exp\left(-\frac{\log(1-x)}{x}\right)=\lim_{n\to +\infty}[x^n](1-x)^{-1/x}\stackrel{?}{=}1.$$
