= Solution
Cancellation of the <factorials> in the consecutive ratio gives
$$
\frac{a_{n+1}}{a_n}=e^{-1}\left(1+\frac1n\right)^{n+1/2},\qquad u_n=\left(n+\frac12\right)\ln\left(1+\frac1n\right)-1.
$$
Set $t=1/n$. The <derivative> of the <natural logarithm> at $1$ gives $\ln(1+t)/t\to1$. Therefore
$$
u_n=\left(1+\frac t2\right)\frac{\ln(1+t)}t-1\longrightarrow\boxed{0}.
$$
Thus the necessary condition from the <term test for divergence> is satisfied. On its own it does not prove that $\sum u_n$ converges.
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