= Solution
The <arithmetic-geometric mean inequality> gives $a_k+1/a_k\ge2$, so $u_n\ge2n\to+\infty$. Hence \b[$(u_n)$ has no finite limit].
For the example, take $\boxed{a_k=1+2^{-k}}$. Every term is positive and different from $1$, and
$$
0<a_k-\frac1{a_k}=\frac{2^{-k}(2+2^{-k})}{1+2^{-k}}<2^{1-k}.
$$
The bounding <geometric series> converges, so the <comparison test for series> proves that the <partial sums> $v_n$ converge.
In general, if $v_n$ converges, the necessary condition for a <convergent series> gives $d_k=a_k-1/a_k=v_k-v_{k-1}\to0$. Since $a_k>0$,
$$
a_k+\frac1{a_k}=\sqrt{\left(a_k-\frac1{a_k}\right)^2+4}=\sqrt{d_k^2+4}\longrightarrow2.
$$
Apply the <Cesaro theorem for convergent sequences> to these summands:
$$
\boxed{\frac{u_n}{n}\longrightarrow2.}
$$
Equivalently, $a_k=(d_k+\sqrt{d_k^2+4})/2\to1$. The <series> of the differences may converge while the <series> of the positive sums still diverges linearly.
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