= Solution
A <rational lower bound for the natural logarithm> proves positivity for every $n$, including $n=1$. Define
$$
h(t)=\ln(1+t)-\frac{2t}{2+t},\qquad t\geq0.
$$
Then $h(0)=0$, while direct <differentiation> gives
$$
h'(t)=\frac1{1+t}-\frac4{(2+t)^2}=\frac{t^2}{(1+t)(2+t)^2}>0\quad(t>0).
$$
Thus $h(t)>0$ for $t>0$. Setting $t=1/n$ in the expression from part (b) gives
$$
u_n=\frac{2+t}{2t}\left[\ln(1+t)-\frac{2t}{2+t}\right]=\frac{2+t}{2t}h(t)>0.
$$
Therefore $\boxed{u_n>0\text{ for every }n\geq1}$, and $(a_n)$ is a strictly increasing <sequence>.
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