= Solution
For $n>1$, the <Taylor series> of the <natural logarithm> converges absolutely at $1/n$:
$$
\ln\left(1+\frac1n\right)=\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{jn^j}.
$$
Multiply this <power series> by $n+1/2$ and subtract $1$. The constant term cancels the subtracted $1$, and the coefficient of $1/n$ is $-1/2+1/2=0$. For $k\geq2$, the coefficient of $n^{-k}$ is
$$
\frac{(-1)^k}{k+1}+\frac{(-1)^{k+1}}{2k}=(-1)^k\left(\frac1{k+1}-\frac1{2k}\right).
$$
Hence
$$
\boxed{u_n=\sum_{k=2}^{\infty}\frac{(-1)^k}{n^k}\left(\frac1{k+1}-\frac1{2k}\right).}
$$
In particular $u_n=1/(12n^2)-1/(12n^3)+O(n^{-4})$, in <Big O notation>.
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